KnowledgeBoat Logo
|
OPEN IN APP

Chapter 3

Expansions

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 3A

Question 1

Using standard formulae, expand each of the following:

(i) (4a + 9)2

(ii) (3x + 10y)2

(iii) (2m+3n)2(\sqrt{2}m + \sqrt{3}n)^2

Answer

We know that,

⇒ (a + b)2 = a2 + b2 + 2ab.

(i) Given,

⇒ (4a + 9)2

⇒ (4a)2 + (9)2 + 2 × 4a × 9

⇒ 16a2 + 81 + 72a.

Hence, (4a + 9)2 = 16a2 + 81 + 72a.

(ii) Given,

⇒ (3x + 10y)2

⇒ (3x)2 + (10y)2 + 2 × 3x × 10y

⇒ 9x2 + 100y2 + 60xy.

Hence, (3x + 10y)2 = 9x2 + 100y2 + 60xy.

(iii) Given,

⇒(2m+3n)2⇒(2m)2+(3n)2+2×2m×3n⇒2m2+3n2+26mn.\Rightarrow (\sqrt{2}m + \sqrt{3}n)^2 \\[1em] \Rightarrow (\sqrt{2}m)^2 + (\sqrt{3}n)^2 + 2 \times \sqrt{2}m \times \sqrt{3}n \\[1em] \Rightarrow 2m^2 + 3n^2 + 2\sqrt{6}mn.

Hence, (2m+3n)2=2m2+3n2+26mn(\sqrt{2}m + \sqrt{3}n)^2 = 2m^2 + 3n^2 + 2\sqrt{6}mn.

Question 2

Using standard formulae, expand each of the following:

(i) (2a2 + 3b)2

(ii) (3x2y + z)2

(iii) (2x+13x)2\Big(2x + \dfrac{1}{3x}\Big)^2

Answer

We know that,

⇒ (a + b)2 = a2 + b2 + 2ab.

(i) Given,

⇒ (2a2 + 3b)2

⇒ (2a2)2 + (3b)2 + 2 × 2a2 × 3b

⇒ 4a4 + 9b2 + 12a2b

Hence, (2a2 + 3b)2 = 4a4 + 9b2 + 12a2b .

(ii) Given,

⇒ (3x2y + z)2

⇒ (3x2y)2 + (z)2 + 2 × 3x2y × z

⇒ 9x4y2 + z2 + 6x2yz

Hence, (3x2y + z)2 = 9x4y2 + z2 + 6x2yz .

(iii) Given,

⇒(2x+13x)2⇒(2x)2+(13x)2+2×2x×13x⇒4x2+19x2+43\Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^2 \\[1em] \Rightarrow (2x)^2 + \Big(\dfrac{1}{3x}\Big)^2 + 2 \times 2x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4x^2 + \dfrac{1}{9x^2} + \dfrac{4}{3}

Hence, (2x+13x)2=4x2+19x2+43\Big(2x + \dfrac{1}{3x}\Big)^2 = 4x^2 + \dfrac{1}{9x^2} + \dfrac{4}{3}.

Question 3

Using standard formulae, expand each of the following:

(i) (25x+56y)2\Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2

(ii) (x3+6x)2\Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2

(iii) (6+5x)2\Big(6 + \dfrac{5}{x}\Big)^2

Answer

We know that,

⇒ (a + b)2 = a2 + b2 + 2ab.

(i) Given,

⇒(25x+56y)2⇒(25x)2+(56y)2+2×25x×56y⇒425x2+2536y2+46xy⇒425x2+2536y2+23xy\Rightarrow \Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{2}{5}x\Big)^2 + \Big(\dfrac{5}{6}y\Big)^2 + 2 \times \dfrac{2}{5}x \times \dfrac{5}{6}y \\[1em] \Rightarrow \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{4}{6}xy \\[1em] \Rightarrow \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{2}{3}xy

Hence, (25x+56y)2=425x2+2536y2+23xy\Big(\dfrac{2}{5}x + \dfrac{5}{6}y\Big)^2 = \dfrac{4}{25}x^2 + \dfrac{25}{36}y^2 + \dfrac{2}{3}xy.

(ii) Given,

⇒(x3+6x)2⇒(x3)2+(6x)2+2×x3×6x⇒x29+36x2+4\Rightarrow \Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{x}{3}\Big)^2 + \Big(\dfrac{6}{x}\Big)^2 + 2 \times \dfrac{x}{3} \times \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2}{9} + \dfrac{36}{x^2} + 4

Hence, (x3+6x)2=x29+36x2+4\Big(\dfrac{x}{3} + \dfrac{6}{x}\Big)^2 = \dfrac{x^2}{9} + \dfrac{36}{x^2} + 4.

(iii) Given,

⇒(6+5x)2⇒(6)2+(5x)2+2×6×5x⇒36+25x2+60x\Rightarrow \Big(6 + \dfrac{5}{x}\Big)^2 \\[1em] \Rightarrow (6)^2 + \Big(\dfrac{5}{x}\Big)^2 + 2 \times 6 \times \dfrac{5}{x} \\[1em] \Rightarrow 36 + \dfrac{25}{x^2} + \dfrac{60}{x}

Hence, (6+5x)2=36+25x2+60x\Big(6 + \dfrac{5}{x}\Big)^2 = 36 + \dfrac{25}{x^2} + \dfrac{60}{x}.

Question 4

Using standard formulae, expand each of the following:

(i) (5x - 3y)2

(ii) (3a - 7b)2

(iii) (12x−32y)2\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2

Answer

We know that,

⇒ (a - b)2 = a2 + b2 - 2ab.

(i) Given,

⇒ (5x - 3y)2

⇒ (5x)2 + (3y)2 - 2 × 5x × 3y

⇒ 25x2 + 9y2 - 30xy

Hence, (5x - 3y)2 = 25x2 + 9y2 - 30xy .

(ii) Given,

⇒ (3a - 7b)2

⇒ (3a)2 + (7b)2 - 2 × 3a × 7b

⇒ 9a2 + 49b2 - 42ab

Hence, (3a - 7b)2 = 9a2 + 49b2 - 42ab.

(iii) Given,

⇒(12x − 32y)2⇒(12x)2+(32y)2−2×12x×32y⇒x24+94y2−32xy\Rightarrow \Big(\dfrac{1}{2}x\ -\ \dfrac{3}{2}y\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{1}{2}x\Big)^2 + \Big(\dfrac{3}{2}y\Big)^2 - 2 \times \dfrac{1}{2}x \times \dfrac{3}{2}y \\[1em] \Rightarrow \dfrac{x^2}{4} + \dfrac{9}{4}y^2 - \dfrac{3}{2}xy

Hence, (12x−32y)2=x24+94y2−32xy\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2 = \dfrac{x^2}{4} + \dfrac{9}{4}y^2 - \dfrac{3}{2}xy.

Question 5

Using standard formulae, expand each of the following:

(i) (a2−b2)2\Big(a^2 - \dfrac{b}{2}\Big)^2

(ii) (3a2b−2b3a)2\Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2

(iii) (5x−23x)2\Big(5x - \dfrac{2}{3x}\Big)^2

Answer

We know that,

⇒ (a - b)2 = a2 + b2 - 2ab.

(i) Given,

⇒(a2−b2)2⇒(a2)2+(b2)2−2×a2×b2⇒a4+b24−a2b\Rightarrow \Big(a^2 - \dfrac{b}{2}\Big)^2 \\[1em] \Rightarrow (a^2)^2 + \Big(\dfrac{b}{2}\Big)^2 - 2 \times a^2 \times \dfrac{b}{2} \\[1em] \Rightarrow a^4 + \dfrac{b^2}{4} - a^2b

Hence, (a2−b2)2=a4+b24−a2b\Big(a^2 - \dfrac{b}{2}\Big)^2 = a^4 + \dfrac{b^2}{4} - a^2b.

(ii) Given,

⇒(3a2b−2b3a)2⇒(3a2b)2+(2b3a)2−2×3a2b×2b3a⇒9a24b2+4b29a2−2\Rightarrow \Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{3a}{2b}\Big)^2 + \Big(\dfrac{2b}{3a}\Big)^2 - 2 \times \dfrac{3a}{2b} \times \dfrac{2b}{3a} \\[1em] \Rightarrow \dfrac{9a^2}{4b^2} + \dfrac{4b^2}{9a^2} - 2

Hence, (3a2b−2b3a)2=9a24b2+4b29a2−2\Big(\dfrac{3a}{2b} - \dfrac{2b}{3a}\Big)^2 = \dfrac{9a^2}{4b^2} + \dfrac{4b^2}{9a^2} - 2.

(iii) Given,

⇒(5x−23x)2⇒(5x)2+(23x)2−2×5x×23x⇒25x2+49x2−203\Rightarrow \Big(5x - \dfrac{2}{3x}\Big)^2 \\[1em] \Rightarrow (5x)^2 + \Big(\dfrac{2}{3x}\Big)^2 - 2 \times 5x \times \dfrac{2}{3x} \\[1em] \Rightarrow 25x^2 + \dfrac{4}{9x^2} - \dfrac{20}{3}

Hence, (5x−23x)2=25x2+49x2−203\Big(5x - \dfrac{2}{3x}\Big)^2 = 25x^2 + \dfrac{4}{9x^2} - \dfrac{20}{3}.

Question 6

Using standard formulae, expand each of the following:

(i) (a + 2b + 3c)2

(ii) (3x + 5y - 2z)2

(iii) (2x - 3y + 7z)2

Answer

We know that,

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).

(i) Given,

⇒ (a + 2b + 3c)2

⇒ (a)2 + (2b)2 + (3c)2 + 2 × (a × 2b + 2b × 3c + 3c × a)

⇒ a2 + 4b2 + 9c2 + 2 × (2ab + 6bc + 3ca)

⇒ a2 + 4b2 + 9c2 + 4ab + 12bc + 6ac

Hence, (a + 2b + 3c)2 = a2 + 4b2 + 9c2 + 4ab + 12bc + 6ac.

(ii) Given,

⇒ (3x + 5y - 2z)2

⇒ [3x + 5y + (-2z)]2

⇒ (3x)2 + (5y)2 + (-2z)2 + 2 × [3x × 5y + 5y × (-2z) + (-2z) × 3x]

⇒ 9x2 + 25y2 + 4z2 + 2 × (15xy - 10yz - 6xz)

⇒ 9x2 + 25y2 + 4z2 + 30xy - 20yz - 12xz

Hence, (3x + 5y - 2z) = 9x2 + 25y2 + 4z2 + 30xy - 20yz - 12xz.

(iii) Given,

⇒ (2x - 3y + 7z)2

⇒ [2x + (-3y) + 7z]2

⇒ (2x)2 + (-3y)2 + (7z)2 + 2 × [2x × (-3y) + (-3y) × (7z) + 7z × 2x]

⇒ 4x2 + 9y2 + 49z2 + 2 × [-6xy - 21yz + 14xz]

⇒ 4x2 + 9y2 + 49z2 - 12xy - 42yz + 28xz.

Hence, (2x - 3y + 7z) = 4x2 + 9y2 + 49z2 - 12xy - 42yz + 28xz.

Question 7

Using standard formulae, expand each of the following:

(i) (6 - 2y + 4z)2

(ii) (4x - 3y + z)2

(iii) (7 - 2x - 3y)2

Answer

We know that,

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).

(i) Given,

⇒ (6 - 2y + 4z)2

⇒ [6 + (-2y) + 4z]2

⇒ (6)2 + (-2y)2 + (4z)2 + 2 × [6 × (-2y) + (-2y) × (4z) + 4z × 6]

⇒ 36 + 4y2 + 16z2 + 2 × [-12y - 8yz + 24z]

⇒ 36 + 4y2 + 16z2 - 24y - 16yz + 48z

Hence, (6 - 2y + 4z)2 = 36 + 4y2 + 16z2 - 24y - 16yz + 48z.

(ii) Given,

⇒ (4x - 3y + z)2

⇒ [4x + (-3y) + z]2

⇒ (4x)2 + (-3y)2 + (z)2 + 2 × [4x × (-3y) + (-3y) × (z) + z × 4x]

⇒ (16x)2 + 9y2 + z2 + 2 × [-12xy - 3yz + 4xz]

⇒ 16x2 + 9y2 + z2 - 24xy - 6yz + 8xz

Hence, (4x - 3y + z)2 = 16x2 + 9y2 + z2 - 24xy - 6yz + 8xz.

(iii) Given,

⇒ (7 - 2x - 3y)2

⇒ [7 + (-2x) + (-3y)]2

⇒ (7)2 + (-2x)2 + (-3y)2 + 2 × [7 × (-2x) + (-2x) × (-3y) + (-3y) × 7]

⇒ 49 + 4x2 + 9y2 + 2 × (-14x + 6xy - 21y)

⇒ 49 + 4x2 + 9y2 - 28x + 12xy - 42y.

Hence, (7 - 2x - 3y)2 = 49 + 4x2 + 9y2 - 28x + 12xy - 42y.

Question 8

Using standard formulae, expand each of the following:

(i) (a2+b3+c4)2\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2

(ii) (2x3+32y−2)2\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2

(iii) (2x+3x−1)2\Big(2x + \dfrac{3}{x} - 1 \Big)^2

Answer

We know that,

⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca).

(i) Given,

⇒(a2+b3+c4)2⇒(a2)2+(b3)2+(c4)2+2×[(a2)×(b3)+(b3)×(c4)+(c4)×(a2)]⇒a24+b29+c216+2×[(ab6)+(bc12)+(ca8)]⇒a24+b29+c216+ab3+bc6+ca4\Rightarrow \Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{2}\Big)^2 +\Big(\dfrac{b}{3}\Big)^2 + \Big(\dfrac{c}{4}\Big)^2 + 2 × \Big[\big(\dfrac{a}{2}\big) × \big(\dfrac{b}{3}\big) + \big(\dfrac{b}{3}\big) × \big(\dfrac{c}{4}\big) + \big(\dfrac{c}{4}\big) × \big(\dfrac{a}{2}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + 2 × \Big[\big(\dfrac{ab}{6}\big) + \big(\dfrac{bc}{12}\big) + \big(\dfrac{ca}{8}\big)\Big] \\[1em] \Rightarrow \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4} \\[1em]

Hence, (a2+b3+c4)2=a24+b29+c216+ab3+bc6+ca4\Big(\dfrac{a}{2} + \dfrac{b}{3} + \dfrac{c}{4}\Big)^2 = \dfrac{a^2}{4} + \dfrac{b^2}{9} + \dfrac{c^2}{16} + \dfrac{ab}{3} + \dfrac{bc}{6} + \dfrac{ca}{4}.

(ii) Given,

⇒(2x3+32y−2)2⇒[2x3+32y+(−2)]2⇒(2x3)2+(32y)2+(−2)2+2×[(2x3)×(32y)+(32y)×(−2)+(−2)×(2x3)]⇒4x29+94y2+4+2×[6x6y−62y−4x3]⇒4x29+94y2+4+2×[xy−3y−4x3]⇒4x29+94y2+4+2xy−6y−8x3\Rightarrow \Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 \\[1em] \Rightarrow \Big[\dfrac{2x}{3} + \dfrac{3}{2y} + (-2)\Big]^2 \\[1em] \Rightarrow \Big(\dfrac{2x}{3}\Big)^2 +\Big(\dfrac{3}{2y}\Big)^2 + (-2)^2 + 2 × \Big[\big(\dfrac{2x}{3}\big) × \big(\dfrac{3}{2y}\big) + \big(\dfrac{3}{2y}\big) × (-2) + (-2) × \big(\dfrac{2x}{3}\big)\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{6x}{6y} - \dfrac{6}{2y} - \dfrac{4x}{3}\Big] \\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + 2 × \Big[\dfrac{x}{y} - \dfrac{3}{y} - \dfrac{4x}{3}\Big]\\[1em] \Rightarrow \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3} \\[1em]

Hence, (2x3+32y−2)2=4x29+94y2+4+2xy−6y−8x3\Big(\dfrac{2x}{3} + \dfrac{3}{2y} - 2\Big)^2 = \dfrac{4x^2}{9} + \dfrac{9}{4y^2} + 4 + \dfrac{2x}{y} - \dfrac{6}{y} - \dfrac{8x}{3}.

(iii) Given,

⇒(2x+3x−1)2⇒[2x+3x+(−1)]2⇒(2x)2+(3x)2+(−1)2+2×[2x×(3x)+(3x)×(−1)+(−1)×(2x)]⇒4x2+9x2+1+2×[6−3x−2x]⇒4x2+9x2+1+12−6x−4x⇒4x2+9x2+13−6x−4x\Rightarrow \Big(2x + \dfrac{3}{x} - 1 \Big)^2 \\[1em] \Rightarrow \Big[2x + \dfrac{3}{x} + (-1) \Big]^2\\[1em] \Rightarrow (2x)^2 +\Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 × \Big[2x × \Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big) × (-1) + (-1) × (2x)\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 2 × \Big[6 - \dfrac{3}{x} - 2x\Big] \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] \Rightarrow 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em]

Hence, (2x+3x−1)2=4x2+9x2+13−6x−4x\Big(2x + \dfrac{3}{x} - 1 \Big)^2 = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x.

Question 9

Using standard formulae, expand each of the following:

(i) (x + 7)(x + 4)

(ii) (a + 13)(a - 8)

(iii) (y - 6)(y - 4)

Answer

(i) Given,

⇒ (x + 7)(x + 4)

⇒ x2 + 4x + 7x + 28

⇒ x2 + 11x + 28.

Hence, (x + 7)(x + 4) = x2 + 11x + 28.

(ii) Given,

⇒ (a + 13)(a - 8)

⇒ a2 - 8a + 13a - 104

⇒ a2 - 5a - 104.

Hence, (a + 13)(a - 8) = a2 + 5a - 104.

(iii) Given,

⇒ (y - 6)(y - 4)

⇒ y2 - 4y - 6y + 24

⇒ y2 - 10y + 24.

Hence, (y - 6)(y - 4) = y2 - 10y + 24.

Question 10

Using standard formulae, expand each of the following:

(i) (9 + 2x)(9 - 3x)

(ii) (5x - 4y)(5x + 3y)

(iii) (3 - 7a)(3 + 4a)

Answer

(i) Given,

⇒ (9 + 2x)(9 - 3x)

⇒ 81 - 27x + 18x - 6x2

⇒ 81 - 9x - 6x2.

Hence, (9 + 2x)(9 - 3x) = 81 - 9x - 6x2.

(ii) Given,

⇒ (5x - 4y)(5x + 3y)

⇒ 25x2 + 15xy - 20xy - 12y2

⇒ 25x2 - 5xy - 12y2.

Hence, (5x - 4y)(5x + 3y) = 25x2 - 5xy - 12y2.

(iii) Given,

⇒ (3 - 7a)(3 + 4a)

⇒ 9 + 12a - 21a - 28a2

⇒ 9 - 9a - 28a2.

Hence, (3 - 7a)(3 + 4a) = 9 - 9a - 28a2.

Question 11

Using standard formulae, expand each of the following:

(i) (3a + 2b)(3a - 2b)

(ii) (5x+15x)(5x−15x)\Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big)

(iii) (2x2+3x2)(2x2−3x2)\Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big)

Answer

We know that,

(a + b)(a - b) = a2 - b2

(i) Given,

⇒ (3a + 2b)(3a - 2b)

⇒ (3a)2 - (2b)2

⇒ 9a2 - 4b2.

Hence, (3a + 2b)(3a - 2b) = 9a2 - 4b2.

(ii) Given,

⇒(5x+15x)(5x−15x)⇒(5x)2−(15x)2⇒25x2−125x2.\Rightarrow \Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big)\\[1em] \Rightarrow (5x)^2 - \Big(\dfrac{1}{5x}\Big)^2 \\[1em] \Rightarrow 25x^2 - \dfrac{1}{25x^2}.

Hence, (5x+15x)(5x−15x)=25x2−125x2\Big(5x + \dfrac{1}{5x}\Big)\Big(5x - \dfrac{1}{5x}\Big) = 25x^2 - \dfrac{1}{25x^2}.

(iii) Given,

⇒(2x2+3x2)(2x2−3x2)⇒(2x2)2−(3x2)2⇒4x4−9x4\Rightarrow \Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big) \\[1em] \Rightarrow (2x^2)^2 - \Big(\dfrac{3}{x^2}\Big)^2 \\[1em] \Rightarrow 4x^4 - \dfrac{9}{x^4} \\[1em]

Hence, (2x2+3x2)(2x2−3x2)=4x4−9x4\Big(2x^2 + \dfrac{3}{x^2}\Big)\Big(2x^2 - \dfrac{3}{x^2}\Big) = 4x^4 - \dfrac{9}{x^4}.

Question 12

Using standard formulae, expand each of the following:

(i) (2 - x)(2 + x)(4 + x2)

(ii) (x + y)(x - y)(x2 + y2)

Answer

We know that,

(a + b)(a - b) = a2 - b2

(i) Given,

⇒ (2 - x)(2 + x)(4 + x2)

⇒ [(2)2 - (x)2](4 + x2)

⇒ (4 - x2)(4 + x2)

⇒ (4)2 - (x2)2

⇒ 16 - x4.

Hence, (2 - x)(2 + x)(4 + x2) = 16 - x4.

(ii) Given,

⇒ (x + y)(x - y)(x2 + y2)

⇒ [(x)2 - (y)2](x2 + y2)

⇒ (x2 - y2)(x2 + y2)

⇒ (x2)2 - (y2)2

⇒ (x4 - y4).

Hence, (x + y)(x - y)(x2 + y2) = (x4 - y4).

Question 13

Using standard formulae, expand each of the following:

(i) (x - 2)(x - 3)(x + 4)

(ii) (x - 5)(2x - 1)(2x + 3)

Answer

(i) Given,

⇒ (x - 2)(x - 3)(x + 4)

⇒ (x2 - 3x - 2x + 6)(x + 4)

⇒ (x2 - 5x + 6)(x + 4)

⇒ x2(x + 4) - 5x(x + 4) + 6(x + 4)

⇒ (x3 + 4x2 - 5x2 - 20x + 6x + 24)

⇒ (x3 - x2 - 14x + 24)

Hence, (x - 2)(x - 3)(x + 4) = x3 - x2 - 14x + 24.

(ii) Given,

⇒ (x - 5)(2x - 1)(2x + 3)

⇒ (2x2 - x - 10x + 5)(2x + 3)

⇒ (2x2 - 11x + 5)(2x + 3)

⇒ 2x2(2x + 3) - 11x(2x + 3) + 5(2x + 3)

⇒ (4x3 + 6x2 - 22x2 - 33x + 10x + 15)

⇒ (4x3 - 16x2 - 23x + 15)

Hence, (x - 5)(2x - 1)(2x + 3) = 4x3 - 16x2 - 23x + 15.

Question 14

Simplify:

(i) (a + b)2 + (a - b)2

(ii) (a + b)2 - (a - b)2

(iii) (x+1x)2+(x−1x)2\Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2

(iv) (x+1x)2−(x−1x)2\Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2

(v) (a2b+2ba)2−(2ba−a2b)2\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2

(vi) (3x−13x)2−(3x+13x)(3x−13x)\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big)

(vii) (5a + 3b)2 - (5a - 3b)2 - 60ab

(viii) (3x + 1)2 - (3x + 2)(3x - 1)

Answer

(i) Given,

⇒ (a + b)2 + (a - b)2

⇒ a2 + b2 + 2ab + a2 + b2 - 2ab

⇒ 2a2 + 2b2

⇒ 2(a2 + b2)

Hence, (a + b)2 + (a - b)2 = 2(a2 + b2).

(ii) Given,

⇒ (a + b)2 - (a - b)2

⇒ (a2 + b2 + 2ab) - (a2 + b2 - 2ab)

⇒ a2 + b2 + 2ab - a2 - b2 + 2ab

⇒ 4ab

Hence, (a + b)2 - (a - b)2 = 4ab.

(iii) Given,

⇒(x+1x)2+(x−1x)2⇒[x2+(1x2)+2×x×(1x)+x2+(1x2)−2×x×(1x)]⇒(x2+1x2+2)+(x2+1x2−2)⇒2x2+2x2⇒2(x2+1x2)\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) + 2 \times x \times \Big(\dfrac{1}{x}\Big) + x^2 + \Big(\dfrac{1}{x^2}\Big) - 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} + 2\Big) + \Big(x^2 + \dfrac{1}{x^2} - 2\Big) \\[1em] \Rightarrow 2x^2 + \dfrac{2}{x^2} \\[1em] \Rightarrow 2\Big(x^2 + \dfrac{1}{x^2}\Big) \\[1em]

Hence, (x+1x)2+(x−1x)2=2(x2+1x2)\Big(x + \dfrac{1}{x}\Big)^2 + \Big(x - \dfrac{1}{x}\Big)^2 = 2\Big(x^2 + \dfrac{1}{x^2}\Big).

(iv) Given,

⇒(x+1x)2−(x−1x)2⇒[x2+(1x2)+2×x×(1x)]−[x2+(1x2)−2×x×(1x)]⇒(x2+1x2+2)−(x2+1x2−2)⇒x2+1x2+2−x2−1x2+2⇒4.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) + 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] - \Big[x^2 + \Big(\dfrac{1}{x^2}\Big) - 2 \times x \times \Big(\dfrac{1}{x}\Big)\Big] \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} + 2\Big) - \Big(x^2 + \dfrac{1}{x^2} - 2\Big) \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} + 2 - x^2 - \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow 4.

Hence, (x+1x)2−(x−1x)2=4\Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4.

(v) Given,

⇒(a2b+2ba)2−(2ba−a2b)2⇒[(a2b)2+(2ba)2+2×a2b×2ba]−[(2ba)2+(a2b)2−2×a2b×2ba]⇒(a24b2+4b2a2+2)−(4b2a2+a24b2−2)⇒a24b2+4b2a2+2−a24b2−4b2a+2⇒4.\Rightarrow \Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2 \\[1em] \Rightarrow \Big[\Big(\dfrac{a}{2b}\Big)^2 + \Big(\dfrac{2b}{a}\Big)^2 + 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] - \Big[\Big(\dfrac{2b}{a}\Big)^2 + \Big(\dfrac{a}{2b}\Big)^2 - 2 \times \dfrac{a}{2b} \times \dfrac{2b}{a}\Big] \\[1em] \Rightarrow \Big(\dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2\Big) - \Big(\dfrac{4b^2}{a^2} + \dfrac{a^2}{4b^2} - 2\Big) \\[1em] \Rightarrow \dfrac{a^2}{4b^2} + \dfrac{4b^2}{a^2} + 2 - \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a} + 2 \\[1em] \Rightarrow 4.

Hence, (a2b+2ba)2−(2ba−a2b)2=4\Big(\dfrac{a}{2b} + \dfrac{2b}{a}\Big)^2 - \Big(\dfrac{2b}{a} - \dfrac{a}{2b}\Big)^2 = 4.

(vi) Given,

⇒(3x−13x)2−(3x+13x)(3x−13x)⇒[(3x)2+(13x)2−2×3x×13x]−[(3x)2−(13x)2]⇒(9x2+19x2−2)−(9x2−19x2)⇒9x2+19x2−2−9x2+19x2⇒29x2−2⇒2(19x2−1)\Rightarrow \Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x}\Big)\Big(3x - \dfrac{1}{3x}\Big) \\[1em] \Rightarrow \Big[(3x)^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times 3x \times \dfrac{1}{3x}\Big] - \Big[(3x)^2 - \Big(\dfrac{1}{3x}\Big)^2\Big] \\[1em] \Rightarrow \Big(9x^2 + \dfrac{1}{9x^2} - 2 \Big) - \Big(9x^2 - \dfrac{1}{9x^2}\Big) \\[1em] \Rightarrow 9x^2 + \dfrac{1}{9x^2} - 2 - 9x^2 + \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{2}{9x^2} - 2 \\[1em] \Rightarrow 2\Big(\dfrac{1}{9x^2} - 1\Big)

Hence, (3x−13x)2−(3x+13x)(3x−13x)=2(19x2−1)\Big(3x - \dfrac{1}{3x}\Big)^2 - \Big(3x + \dfrac{1}{3x})\Big(3x - \dfrac{1}{3x}\Big) = 2\Big(\dfrac{1}{9x^2} - 1\Big).

(vii) Given,

⇒ (5a + 3b)2 - (5a - 3b)2 - 60ab

⇒ [(5a)2 + (3b)2 + 2 × 5a × 3b] - [(5a)2 + (3b)2 - 2 × 5a × 3b] - 60ab

⇒ [25a2 + 9b2 + 2 × 5a × 3b] - [25a2 + 9b2 - 2 × 5a × 3b] - 60ab

⇒ 25a2 + 9b2 + 30ab - 25a2 - 9b2 + 30ab - 60ab

⇒ 60ab - 60ab

⇒ 0

Hence, (5a + 3b)2 - (5a - 3b)2 - 60ab = 0.

(viii) Given,

⇒ (3x + 1)2 - [(3x + 2)(3x - 1)]

⇒ (3x)2 + (1)2 + 2 × 3x × 1 - (9x2 - 3x + 6x - 2)

⇒ 9x2 + 1 + 6x - (9x2 + 3x - 2)

⇒ 9x2 + 1 + 6x - 9x2 - 3x + 2

⇒ 1 + 3x + 2

⇒ 3x + 3

⇒ 3(x + 1).

Hence, (3x + 1)2 - (3x + 2)(3x - 1) = 3(x + 1).

Question 15

(i) If (a + b) = 7 and ab = 10, find the value of (a - b).

(ii) If (x - y) = 5 and xy = 24, find the value of (x + y).

Answer

(i) Given,

(a + b) = 7 and ab = 10

Using identity,

⇒ (a + b)2 - (a - b)2 = 4ab

Substituting values we get :

⇒ (7)2 - (a - b)2 = 4 × 10

⇒ 49 - (a - b)2 = 40

⇒ (a - b)2 = 49 - 40

⇒ (a - b)2 = 9

⇒ (a - b)2 = 9\sqrt{9}

⇒ (a - b) = ±3\pm 3

Hence, (a - b) = ±3\pm 3.

(ii) Given,

(x - y) = 5 and xy = 24

Using identity,

⇒ (x + y)2 - (x - y)2 = 4xy

⇒ (x + y)2 = 4xy + (x - y)2

⇒ (x + y)2 = 4 × 24 + (5)2

⇒ (x + y)2 = 96 + 25

⇒ (x + y) = 121\sqrt{121}

⇒ (x + y) = ±11\pm 11

Hence, (x + y) = ±11\pm 11.

Question 16

If (3a + 4b) = 16 and ab = 4, find the value of (9a2 + 16b2).

Answer

⇒ (3a + 4b)2 = (3a)2 + (4b)2 + 2 × 3a × 4b

⇒ (3a + 4b)2 = 9a2 + 16b2 + 24ab

⇒ 9a2 + 16b2 = (3a + 4b)2 - 24ab

Given,

(3a + 4b) = 16 and ab = 4

Substituting values we get :

⇒ 9a2 + 16b2 = (16)2 - 24 × 4

⇒ 9a2 + 16b2 = 256 - 96

⇒ 9a2 + 16b2 = 160.

Hence, 9a2 + 16b2 = 160.

Question 17

If (a + b) = 2 and (a - b) = 10, find the values of :

(i) (a2 + b2)

(ii) ab

Answer

(i) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)2 + (a - b)2 = 2(a2 + b2)

⇒ (2)2 + (10)2 = 2(a2 + b2)

⇒ 2(a2 + b2) = 4 + 100

⇒ 2(a2 + b2) = 104

⇒ a2 + b2 = 52.

Hence, a2 + b2 = 52.

(ii) Given,

(a + b) = 2 and (a - b) = 10

Using identity,

⇒ (a + b)2 - (a - b)2 = 4ab

⇒ (2)2 - (10)2 = 4ab

⇒ 4ab = 4 - 100

⇒ 4ab = -96

⇒ ab = -24

Hence, ab = -24.

Question 18

If (a - b) = 0.9 and ab = 0.36, find the values of :

(i) (a + b).

(ii) (a2 - b2).

Answer

(i) Given,

(a - b) = 0.9 and ab = 0.36

Using identity,

⇒ (a + b)2 - (a - b)2 = 4ab

⇒ (a + b)2 = 4ab + (a - b)2

⇒ (a + b)2 = 4 × 0.36 + (0.9)2

⇒ (a + b)2 = 1.44 + 0.81

⇒ (a + b)2 = 2.25

⇒ (a + b) = 2.25\sqrt{2.25}

⇒ (a + b) = ±1.5\pm 1.5

Hence, (a + b) = ±1.5\pm 1.5

(ii) Using identity,

⇒ a2 − b2 = (a − b)(a + b)

⇒ a2 − b2 = 0.9 × ±1.5\pm 1.5

⇒ a2 − b2 = ±1.35\pm 1.35

Hence, a2 − b2 = ±1.35\pm 1.35

Question 19

If (x+1x)=5\Big(x + \dfrac{1}{x}\Big) = 5, find the values of :

(i) (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big)

(ii) (x4+1x4)\Big(x^4 + \dfrac{1}{x^4}\Big)

Answer

(i) Given,

(x+1x)=5\Big(x + \dfrac{1}{x}\Big) = 5

⇒(x+1x)2=x2+(1x)2+2×x×1x⇒(5)2=x2+1x2+2×x×1x⇒25=x2+1x2+2⇒x2+1x2=25−2⇒x2+1x2=23.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (5)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 25 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 25 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 23.

Hence, x2+1x2=23x^2 + \dfrac{1}{x^2} = 23.

(ii) Given,

(x+1x)=5\Big(x + \dfrac{1}{x}\Big) = 5

From part (i),

x2+1x2=23x^2 + \dfrac{1}{x^2} = 23

Using identity,

⇒(x2+1x2)2=(x2)2+(1x2)2+2×x2×1x2⇒(23)2=(x2)2+(1x2)2+2×x2×1x2⇒529=x4+1x4+2⇒x4+1x4=529−2⇒x4+1x4=527\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (23)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 529 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 529 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 527 \\[1em]

Hence, x4+1x4=527x^4 + \dfrac{1}{x^4} = 527.

Question 20

If (x−1x)=4\Big(x - \dfrac{1}{x}\Big) = 4, find the values of :

(i) (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big)

(ii) (x4+1x4)\Big(x^4 + \dfrac{1}{x^4}\Big).

Answer

(i) Given,

(x−1x)=4\Big(x - \dfrac{1}{x}\Big) = 4

⇒(x−1x)2=x2+(1x)2−2×x×1x⇒(4)2=x2+(1x)2−2×x×1x⇒16=x2+1x2−2⇒x2+1x2=16+2⇒x2+1x2=18\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (4)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 16 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 16 + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 18

Hence, x2+1x2=18x^2 + \dfrac{1}{x^2} = 18.

(ii) Given,

(x−1x)=4\Big(x - \dfrac{1}{x}\Big) = 4

From part (i),

x2+1x2=18x^2 + \dfrac{1}{x^2} = 18

⇒(x2+1x2)2=(x2)2+(1x2)2+2×x2×1x2⇒(18)2=(x2)2+(1x2)2+2×x2×1x2⇒324=x4+1x4+2⇒x4+1x4=324−2⇒x4+1x4=322.\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (18)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 324 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 324 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 322.

Hence, x4+1x4=322x^4 + \dfrac{1}{x^4} = 322.

Question 21

If x−2=13xx - 2 = \dfrac{1}{3x}, find the values of :

(i) (x2+19x2)\Big(x^2 + \dfrac{1}{9x^2}\Big)

(ii) (x4+181x4)\Big(x^4 + \dfrac{1}{81x^4}\Big).

Answer

(i) Given,

⇒x−2=13x⇒x−13x=2\Rightarrow x - 2 = \dfrac{1}{3x} \\[1em] \Rightarrow x - \dfrac{1}{3x} = 2

We know that,

⇒(x−13x)2=x2+(13x)2−2×x×13x⇒(2)2=x2+(13x)2−2×x×13x⇒4=x2+19x2−23⇒x2+19x2=4+23⇒x2+19x2=12+23⇒x2+19x2=143\Rightarrow \Big(x - \dfrac{1}{3x}\Big)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow (2)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{9x^2} - \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = 4 + \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{12 + 2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}

Hence, x2+19x2=143x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}.

(ii) From part (i),

x2+19x2=143x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}

Using identity,

⇒(x2+19x2)2=(x2)2+(19x2)2+2×x2×19x2⇒(143)2=(x2)2+(19x2)2+2×x2×19x2⇒1969=x4+181x4+29⇒x4+181x4=1969−29⇒x4+181x4=196−29⇒x4+181x4=1949\Rightarrow \Big(x^2 + \dfrac{1}{9x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \Big(\dfrac{14}{3}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{196}{9} = x^4 + \dfrac{1}{81x^4} + \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196}{9} - \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196 - 2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}

Hence, x4+181x4=1949x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}.

Question 22

If (x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6, find the values of :

(i) (x−1x)\Big(x - \dfrac{1}{x}\Big).

(ii) (x2−1x2)\Big(x^2 - \dfrac{1}{x^2}\Big)

Answer

(i) Given,

(x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6

We know that,

⇒(x+1x)2−(x−1x)2=4⇒(6)2−(x−1x)2=4⇒36−4=(x−1x)2⇒32=(x−1x)2⇒(x−1x)=32⇒(x−1x)=±42.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (6)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 36 - 4 = \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow 32 = \Big(x - \dfrac{1}{x}\Big)^2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{32} \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 4\sqrt{2}.

Hence, (x−1x)=±42\Big(x - \dfrac{1}{x}\Big) = \pm 4\sqrt{2}.

(ii) Given,

(x+1x)=6\Big(x + \dfrac{1}{x}\Big) = 6

From part (i),

⇒(x−1x)=±42\Rightarrow \Big(x - \dfrac{1}{x}\Big) = ±4\sqrt{2}

We know that,

⇒(x2−1x2)=(x+1x)(x−1x)⇒(x2−1x2)=6×±42⇒(x2−1x2)=±242\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = 6 \times \pm 4\sqrt{2} \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 24\sqrt{2}

Hence, x2−1x2=±242x^2 - \dfrac{1}{x^2} = \pm 24\sqrt{2}.

Question 23

If (x−1x)=8\Big(x - \dfrac{1}{x}\Big) = 8, find the values of :

(i) (x+1x)\Big(x + \dfrac{1}{x}\Big)

(ii) (x2−1x2)\Big(x^2 - \dfrac{1}{x^2}\Big)

Answer

(i) Given,

(x−1x)=8\Big(x - \dfrac{1}{x}\Big) = 8

We know that,

⇒(x+1x)2−(x−1x)2=4⇒(x+1x)2−(8)2=4⇒(x+1x)2−64=4⇒(x+1x)2=64+4⇒(x+1x)2=68⇒(x+1x)=68⇒(x+1x)=±217\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - (8)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 64 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 68 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{68} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17} \\[1em]

Hence, (x+1x)=±217\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}.

(ii) Given,

(x−1x)=8\Big(x - \dfrac{1}{x}\Big) = 8

From (i),

(x+1x)=±217\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}

Using identity,

⇒(x2−1x2)=(x+1x)(x−1x)⇒(x2−1x2)=(±217)×8⇒(x2−1x2)=±1617\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 2\sqrt{17})\times 8 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 16\sqrt{17}

Hence, x2−1x2=±1617x^2 - \dfrac{1}{x^2} = \pm 16\sqrt{17}.

Question 24

If (x2+1x2)=7\Big(x^2 + \dfrac{1}{x^2}\Big) = 7, find the values of :

(i) (x+1x)\Big(x + \dfrac{1}{x}\Big)

(ii) (x−1x)\Big(x - \dfrac{1}{x}\Big)

(iii) (2x2−2x2)\Big(2x^2 - \dfrac{2}{x^2}\Big).

Answer

(i) Given,

(x2+1x2)=7\Big(x^2 + \dfrac{1}{x^2}\Big) = 7

Using identity,

⇒(x+1x)2=x2+1x2+2⇒(x+1x)2=7+2⇒(x+1x)2=9⇒(x+1x)=±9⇒(x+1x)=±3.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 7 + 2 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 3.

Hence, (x+1x)=±3\Big(x + \dfrac{1}{x}\Big) = \pm 3.

(ii) Given,

(x2+1x2)=7\Big(x^2 + \dfrac{1}{x^2}\Big) = 7

Using identity,

⇒(x−1x)2=x2+1x2−2⇒(x−1x)2=7−2⇒(x−1x)2=5⇒(x−1x)=±5\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 7 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 5 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5} \\[1em]

Hence, (x−1x)=±5\Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}.

(iii) Given,

(x2+1x2)=7\Big(x^2 + \dfrac{1}{x^2}\Big) = 7

From part (i) and (ii),

(x+1x)=±3 and (x−1x)=±5\Big(x + \dfrac{1}{x}\Big) = \pm 3 \text{ and } \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{5}

Using identity,

⇒(x2−1x2)=(x+1x)(x−1x)⇒(x2−1x2)=(±3)×(±5)⇒(x2−1x2)=±35⇒2(x2−1x2)=2×±35⇒(2x2−2x2)=2×(±35)⇒(2x2−2x2)=±65\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 3) \times (\pm \sqrt{5}) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 3\sqrt{5} \\[1em] \Rightarrow 2\Big(x^2 - \dfrac{1}{x^2}\Big) = 2 \times \pm 3\sqrt{5} \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = 2 \times (\pm 3\sqrt{5}) \\[1em] \Rightarrow \Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}

Hence, (2x2−2x2)=±65\Big(2x^2 - \dfrac{2}{x^2}\Big) = \pm 6\sqrt{5}.

Question 25

If (x2+125x2)=925\Big(x^2 + \dfrac{1}{25x^2}\Big) = 9\dfrac{2}{5}, find the value of (x−15x)\Big(x - \dfrac{1}{5x}\Big).

Answer

⇒(x−15x)2=[x2+(15x)2−2×x×15x]⇒(x−15x)2=[x2+125x2−25]⇒(x−15x)2=925−25⇒(x−15x)2=475−25⇒(x−15x)2=47−25⇒(x−15x)2=455⇒(x−15x)2=9⇒(x−15x)=9⇒(x−15x)=±3\Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \Big(\dfrac{1}{5x}\Big)^2 - 2 \times x \times \dfrac{1}{5x}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \Big[x^2 + \dfrac{1}{25x^2} - \dfrac{2}{5}\Big] \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9\dfrac{2}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47}{5} - \dfrac{2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{47 - 2}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow (x - \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x - \dfrac{1}{5x}\Big) = \pm 3 \\[1em]

Hence, (x−15x)=±3\Big(x - \dfrac{1}{5x}\Big) = \pm 3.

Question 26

If a2−4a−1=0a^2 - 4a - 1 = 0 and a≠0a \neq 0, find the values of:

(i) (a−1a)\Big(a - \dfrac{1}{a}\Big)

(ii) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(iii) (a2−1a2)\Big(a^2 - \dfrac{1}{a^2}\Big)

(iv) (a2+1a2)\Big(a^2 + \dfrac{1}{a^2}\Big)

Answer

(i) Solving,

⇒a2−4a−1=0⇒a2−4a=1⇒a(a−4)=1⇒a−4=1a⇒a−1a=4\Rightarrow a^2 - 4a - 1 = 0 \\[1em] \Rightarrow a^2 - 4a = 1 \\[1em] \Rightarrow a(a - 4) = 1 \\[1em] \Rightarrow a - 4 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 4 \\[1em]

Hence, (a−1a)=4\Big(a - \dfrac{1}{a}\Big) = 4

(ii) Using identity,

⇒(a+1a)2−(a−1a)2=4⇒(a+1a)2−(4)2=4⇒(a+1a)2−16=4⇒(a+1a)2=4+16⇒(a+1a)2=20⇒a+1a=20⇒a+1a=±25.\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (4)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 16 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 16\\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 20 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{20} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}.

Hence, a+1a=±25a + \dfrac{1}{a} = \pm 2\sqrt{5}

(iii) From part (i) and (ii),

⇒a−1a=4⇒a+1a=±25\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}

Case 1:

⇒a+1a=25\Rightarrow a + \dfrac{1}{a} = 2\sqrt{5}

Using identity,

⇒(a+1a)(a−1a)=(a2−1a2)⇒(25)×(4)=(a2−1a2)⇒(a2−1a2)=85\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = 8\sqrt{5} \\[1em]

Case 2:

⇒a+1a=−25\Rightarrow a + \dfrac{1}{a} = -2\sqrt{5}

Using identity,

⇒(a+1a)(a−1a)=(a2−1a2)⇒(−25)×(4)=(a2−1a2)⇒(a2−1a2)=−85\Rightarrow \Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (-2\sqrt{5}) \times (4) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = -8\sqrt{5} \\[1em]

Hence, (a2−1a2)=±85\Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 8\sqrt{5}.

(iv) From part (i) and (ii),

⇒a−1a=4⇒a+1a=±25\Rightarrow a - \dfrac{1}{a} = 4 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 2\sqrt{5}

Using identity,

⇒(a+1a)2+(a−1a)2=2(a2+1a2)⇒(±25)2+(4)2=2(a2+1a2)⇒2(a2+1a2)=20+16⇒2(a2+1a2)=36⇒(a2+1a2)=362⇒(a2+1a2)=18\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow (\pm 2\sqrt{5})^2 + (4)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 20 + 16 \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 36 \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = \dfrac{36}{2} \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = 18 \\[1em]

Hence, (a2+1a2)=18\Big(a^2 + \dfrac{1}{a^2}\Big) = 18.

Question 27

If a=1a−5a = \dfrac{1}{a - 5}, where a≠5 and a≠0a \neq 5 \text{ and }a \neq 0, find the values of:

(i) (a−1a)\Big(a - \dfrac{1}{a}\Big)

(ii) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(iii) (a2−1a2)\Big(a^2 - \dfrac{1}{a^2}\Big)

(iv) (a2+1a2)\Big(a^2 + \dfrac{1}{a^2}\Big)

Answer

(i) Given,

a=1a−5a = \dfrac{1}{a - 5}

⇒a=1a−5⇒a−5=1a⇒a−1a=5\Rightarrow a = \dfrac{1}{a - 5} \\[1em] \Rightarrow a - 5 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 5

Hence, (a−1a)=5\Big(a - \dfrac{1}{a}\Big) = 5

(ii) From part (i),

a−1a=5a - \dfrac{1}{a} = 5

Using identity,

⇒(a+1a)2−(a−1a)2=4⇒(a+1a)2−(5)2=4⇒(a+1a)2=4+25⇒(a+1a)2=29⇒a+1a=±29\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (5)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 25\\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 29 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}

Hence, a+1a=±29a + \dfrac{1}{a} = \pm \sqrt{29}

(iii) From (i) and (ii),

⇒a−1a=5⇒a+1a=±29\Rightarrow a - \dfrac{1}{a} = 5 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}

Using identity,

(a+1a)(a−1a)=(a2−1a2)\Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big)

⇒(±29)×(5)=(a2−1a2)⇒(a2−1a2)=±529\Rightarrow (\pm \sqrt{29}) \times (5) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 5\sqrt{29} \\[1em]

Hence, (a2−1a2)=±529\Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 5\sqrt{29}

(iv) From (i) and (ii),

⇒a−1a=5⇒a+1a=±29\Rightarrow a - \dfrac{1}{a} = 5 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm\sqrt{29}

Using identity,

(a+1a)2+(a−1a)2=2(a2+1a2)\Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big)

⇒(±29)2+(5)2=2(a2+1a2)⇒2(a2+1a2)=29+25⇒2(a2+1a2)=54⇒(a2+1a2)=542⇒(a2+1a2)=27\Rightarrow (\pm \sqrt{29})^2 + (5)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 29 + 25 \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 54 \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = \dfrac{54}{2} \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = 27 \\[1em]

Hence, (a2+1a2)=27\Big(a^2 + \dfrac{1}{a^2}\Big) = 27.

Question 28

Using (a + b)2 = (a2 + b2 + 2ab), evaluate:

(i) (137)2

(ii) (1008)2

(iii) (11.6)2

Answer

(i) Given,

⇒ (137)2

⇒ (130 + 7)2

Using identity :

(a + b)2 = a2 + b2 + 2ab

⇒ (130 + 7)2 = (130)2 + 72 + 2 × 130 × 7

⇒ (130 + 7)2 = 16900 + 49 + 1820 = 18769.

Hence, (137)2 = 18769.

(ii) Given,

⇒ (1008)2

⇒ (1000 + 8)2

Using identity :

(a + b)2 = a2 + b2 + 2ab

⇒ (1000 + 8)2 = (1000)2 + 82 + 2 × 1000 × 8

⇒ (1000 + 8)2 = 1000000 + 64 + 16000 = 1016064.

Hence, (1008)2 = 1016064.

(iii) Given,

⇒ (11.6)2

⇒ (11 + 0.6)2

Using identity :

(a + b)2 = a2 + b2 + 2ab

⇒ (11 + 0.6)2 = (11)2 + (0.6)2 + 2 × 11 × 0.6

⇒ (11 + 0.6)2 = 121 + 0.36 + 13.2 = 134.56

Hence, (11.6)2 = 134.56.

Question 29

Using (a - b)2 = (a2 + b2 - 2ab), evaluate:

(i) (97)2

(ii) (992)2

(iii) (9.98)2

Answer

(i) Given,

⇒ (97)2

⇒ (100 - 3)2

Using identity :

⇒ (a - b)2 = a2 + b2 - 2ab

⇒ (100 - 3)2 = (100)2 + 32 - 2 × 100 × 3

⇒ (100 - 3)2 = 10000 + 9 - 600

⇒ 9409.

Hence, (97)2 = 9409.

(ii) Given,

⇒ (992)2

⇒ (1000 - 8)2

Using identity :

(a - b)2 = a2 + b2 - 2ab

⇒ (1000 - 8)2 = (1000)2 + 82 - 2 × 1000 × 8

⇒ (1000 - 8)2 = 1000000 + 64 - 16000

⇒ 984064.

Hence, (992)2 = 984064.

(iii) Given,

⇒ (9.98)2

⇒ (10 - 0.02)2

Using identity :

(a - b)2 = a2 + b2 - 2ab

⇒ (10 - 0.02)2 = (10)2 + 0.022 - 2 × 10 × 0.02

⇒ (10 - 0.02)2 = 100 + 0.0004 - 0.4

⇒ 99.6004

Hence, (9.98)2 = 99.6004.

Question 30

Fill in the blanks to make the given expression a perfect square:

(i) 16a2 + 9b2 + ..............

(ii) 25a2 + 16b2 - ..............

(iii) 4a2 + 20ab + ..............

(iv) 9a2 - 24ab + ..............

Answer

(i) Given,

16a2 + 9b2 + ..............

Adding 24ab to above equation, we get :

⇒ 16a2 + 9b2 + 24ab

⇒ (4a)2 + (3b)2 + 2 × 4a × 3b

We know that,

(a + b)2 = a2 + b2 + 2ab

⇒ (4a + 3b)2.

Hence, on adding 24ab to the expression 16a2 + 9b2, it becomes a perfect square.

(ii) Given,

25a2 + 16b2 + ..............

Adding 40ab to above equation, we get :

⇒ 25a2 + 16b2 + 40ab

⇒ (5a)2 + (4b)2 + 2 × 5a × 4b

We know that,

(a + b)2 = a2 + b2 + 2ab

⇒ (5a + 4b)2.

Hence, on adding 40ab to the expression 25a2 + 16b2, it becomes a perfect square.

(iii) Given,

4a2 + 20ab + ..............

Adding 25b2 to above equation, we get :

⇒ 4a2 + 25b2 + 20ab

⇒ (2a)2 + (5b)2 + 2 × 2a × 5b

We know that,

(a + b)2 = a2 + b2 + 2ab

⇒ (2a + 5b)2

Hence, on adding 25b2 to the expression 4a2 + 20ab, it becomes a perfect square.

(iv) Given,

9a2 - 24ab + ..............

Adding 16b2 to above equation, we get :

⇒ 9a2 + 16b2 - 24ab

⇒ (3a)2 + (4b)2 - 2 × 3a × 4b

We know that,

(a - b)2 = a2 + b2 - 2ab

⇒ (3a - 4b)2

Hence, on adding 16b2 to the expression 9a2 - 24ab, it becomes a perfect square.

Question 31

If (a + b + c) = 14 and (a2 + b2 + c2) = 74, find the value of (ab + bc + ca).

Answer

Given,

(a + b + c) = 14

(a2 + b2 + c2) = 74

Using identity,

⇒ (a + b + c)2 = (a2 + b2 + c2) + 2 (ab + bc + ca)

⇒ (14)2 = (74) + 2 (ab + bc + ca)

⇒ 196 - 74 = 2 (ab + bc + ca)

⇒ 2 (ab + bc + ca) = 122

⇒ (ab + bc + ca) = 1222\dfrac{122}{2}

⇒ (ab + bc + ca) = 61.

Hence, (ab + bc + ca) = 61.

Question 32

If (a + b + c) = 15 and (ab + bc + ca) = 74, find the value of (a2 + b2 + c2).

Answer

Given,

(a + b + c) = 15

(ab + bc + ca) = 74

Using identity,

⇒ (a + b + c)2 = (a2 + b2 + c2) + 2 (ab + bc + ca)

⇒ (15)2 = (a2 + b2 + c2) + 2 (74)

⇒ 225 = (a2 + b2 + c2) + 148

⇒ 225 - 148 = (a2 + b2 + c2)

⇒ (a2 + b2 + c2) = 77

Hence, (a2 + b2 + c2) = 77.

Question 33

If (a2 + b2 + c2) = 50 and (ab + bc + ca) = 47, find the value of (a + b + c).

Answer

Given,

(a2 + b2 + c2) = 50

(ab + bc + ca) = 47

Using identity,

⇒ (a + b + c)2 = (a2 + b2 + c2) + 2 (ab + bc + ca)

⇒ (a + b + c)2 = (50) + 2 (47)

⇒ (a + b + c)2 = 50 + 94

⇒ (a + b + c)2 = 144

⇒ (a + b + c) = 144\sqrt{144}

⇒ (a + b + c) = ±12\pm 12

Hence, (a + b + c) = ±12\pm 12.

Question 34

If (a2 + b2 + c2) = 89 and (ab - bc - ca) = 16, find the value of (a + b - c).

Answer

Given,

(a2 + b2 + c2) = 89

(ab - bc - ca) = 16

Using identity,

⇒ (a + b - c)2 = (a2 + b2 + c2) + 2 (ab - bc - ca)

⇒ (a + b - c)2 = (89) + 2 (16)

⇒ (a + b - c)2 = 89 + 32

⇒ (a + b - c)2 = 121

⇒ (a + b - c) = 121\sqrt{121}

⇒ (a + b - c) = ±11\pm 11

Hence, (a + b - c) = ±11\pm 11.

Exercise 3B

Question 1

Expand:

(i) (3a + 5b)3

(ii) (2p - 3q)3

(iii) (2x+13x)3\Big(2x + \dfrac{1}{3x}\Big)^3

(iv) (3ab - 2c)3

(v) (3a−1a)3\Big(3a - \dfrac{1}{a}\Big)^3

(vi) (12x−23y)3\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3

Answer

(i) Given,

(3a + 5b)3

Using identity :

(a + b)3 = a3 + b3 + 3ab(a + b)

⇒ (3a + 5b)3 = (3a)3 + (5b)3 + 3 × 3a × 5b × (3a + 5b)

⇒ (3a + 5b)3 = 27a3 + 125b3 + 45ab × (3a + 5b)

⇒ (3a + 5b)3 = 27a3 + 135a2 b + 225ab2 + 125b3

Hence, (3a + 5b)3 = 27a3 + 135a2b + 225ab2 + 125b3.

(ii) Given,

(2p - 3q)3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒ (2p - 3q)3 = [(2p)3 - (3q)3 - 3 × (2p)2 (3q) + 3 × (2p) × (3q)2]

⇒ (2p - 3q)3 = 8p3 - 27q3 - 3 × 4p2 (3q) + 3 × (2p) × 9q2

⇒ (2p - 3q)3 = 8p3 - 36p2q + 54pq2 - 27q3

Hence, (2p - 3q)3 = 8p3 - 36p2q + 54pq2 - 27q3.

(iii) Given,

(2x+13x)3\Big(2x + \dfrac{1}{3x}\Big)^3

Using identity :

(a + b)3 = a3 + b3 + 3a2b + 3ab2

⇒(2x+13x)3=[(2x)3+(13x)3+3×(2x)2×(13x)+3×2x×(13x)2]⇒(2x+13x)3=8x3+127x3+3×4x2×(13x)+3×2x×(19x2)⇒(2x+13x)3=8x3+4x+23x+127x3\Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = \Big[(2x)^3 + \Big(\dfrac{1}{3x}\Big)^3 + 3 \times (2x)^2 \times \Big(\dfrac{1}{3x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{3x}\Big)^2 \Big]\\[1em] \Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + \dfrac{1}{27x^3} + 3 \times 4x^2 \times \Big(\dfrac{1}{3x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{9x^2}\Big) \\[1em] \Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + 4x + \dfrac{2}{3x} + \dfrac{1}{27x^3} \\[1em]

Hence, (2x+13x)3=8x3+4x+23x+127x3\Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + 4x + \dfrac{2}{3x} + \dfrac{1}{27x^3}.

(iv) Given,

(3ab - 2c)3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒ (3ab - 2c)3 = [(3ab)3 - (2c)3 - 3 × (3ab)2 (2c) + 3 × (3ab) × (2c)2]

⇒ (3ab - 2c)3 = 27a3b3 - 8c3 - (9a2b2) × 6c + 3 × (3ab) × 4c2

⇒ (3ab - 2c)3 = 27a3b3 - 54a2b2c + 36 abc2 - 8c3

Hence, (3ab - 2c)3 = 27a3b3 - 54a2b2c + 36 abc2 - 8c3.

(v) Given,

(3a−1a)3\Big(3a - \dfrac{1}{a}\Big)^3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒(3a−1a)3=[(3a)3−(1a)3−3×(3a)2×(1a)+3×3a×(1a)2]⇒(3a−1a)3=27a3−1a3−3×9a2×(1a)+3×3a×(1a2)⇒(3a−1a)3=27a3−27a+9a−1a3\Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = \Big[(3a)^3 - \Big(\dfrac{1}{a}\Big)^3 - 3 \times (3a)^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times 3a \times \Big(\dfrac{1}{a}\Big)^2 \Big]\\[1em] \Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - \dfrac{1}{a^3} - 3 \times 9a^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times 3a \times \Big(\dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - 27a + \dfrac{9}{a} - \dfrac{1}{a^3} \\[1em]

Hence, (3a−1a)3=27a3−27a+9a−1a3\Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - 27a + \dfrac{9}{a} - \dfrac{1}{a^3}.

(vi) Given,

(12x−23y)3\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒(12x−23y)3=[(12x)3−(23y)3−3×(12x)2×(23y)+3×(12x)×(23y)2]⇒(12x−23y)3=(18x3)−(827y3)−3×(14x2)×(23y)+3×(12x)×(49y2)⇒(12x−23y)3=18x3−827y3−612x2y+1218xy2⇒(12x−23y)3=18x3−827y3−12x2y+23xy2\Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \Big[\Big(\dfrac{1}{2}x\Big)^3 - \Big(\dfrac{2}{3}y\Big)^3 - 3 \times \Big(\dfrac{1}{2}x\Big)^2 \times \Big(\dfrac{2}{3}y\Big) + 3 \times \Big(\dfrac{1}{2}x\Big) \times \Big(\dfrac{2}{3}y\Big)^2 \Big]\\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \Big(\dfrac{1}{8}x^3\Big) - \Big(\dfrac{8}{27}y^3\Big) - 3 \times \Big(\dfrac{1}{4}x^2\Big) \times \Big(\dfrac{2}{3}y\Big) + 3 \times \Big(\dfrac{1}{2}x\Big) \times \Big(\dfrac{4}{9}y^2\Big) \\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{6}{12}x^2y + \dfrac{12}{18}xy^2 \\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{1}{2}x^2y + \dfrac{2}{3}xy^2 \\[1em]

Hence, (12x−23y)3=18x3−827y3−12x2y+23xy2\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{1}{2}x^2y + \dfrac{2}{3}xy^2.

Question 2

If 4a + 3b = 10 and ab = 2, find the value of 64a3+27b364a^3 + 27b^3.

Answer

Given,

(4a + 3b) = 10

ab = 2

⇒ (4a + 3b)3 = [(4a)3 + (3b)3 + 3 × 4a × 3b × (4a + 3b)]

⇒ (10)3 = 64a3 + 27b3 + 36ab × (10)

⇒ 1000 = 64a3 + 27b3 + 36 × 2 × 10

⇒ 1000 = 64a3 + 27b3 + 720

⇒ 64a3 + 27b3 = 1000 - 720

⇒ 64a3 + 27b3 = 280

Hence, 64a3 + 27b3 = 280.

Question 3

If 3x – 2y = 5 and xy = 6, find the value of 27x3−8y327x^3 - 8y^3.

Answer

Given,

3x – 2y = 5

xy = 6

Using identity :

(a - b)3 = a3 - b3 - 3ab(a - b)

⇒ (3x – 2y)3 = [(3x)3 - (2y)3 - 3 × 3x × 2y × (3x - 2y)]

⇒ (5)3 = 27x3 - 8y3 - 18xy × (5)

⇒ 125 = 27x3 - 8y3 - 90xy

⇒ 125 = 27x3 - 8y3 - 90(6)

⇒ 27x3 - 8y3 = 125 + 540

⇒ 27x3 - 8y3 = 665

Hence, 27x3 - 8y3 = 665.

Question 4

If a + 3b = 6, show that a3+27b3+54ab=216a^3 + 27b^3 + 54ab = 216.

Answer

Given,

a + 3b = 6

⇒ (a + 3b)3 = [(a)3 + (3b)3 + 3 × a × 3b × (a + 3b)]

⇒ (6)3 = a3 + 27b3 + 9ab × (a + 3b)

⇒ 216 = a3 + 27b3 + 9ab × (6)

⇒ a3 + 27b3 + 54ab = 216

Hence proved that a3 + 27b3 + 54ab = 216.

Question 5

If a + 2b + 3c = 0, show that a3 + 8b3 + 27c3 = 18abc.

Answer

We know that,

If x + y + z = 0 then x3 + y3 + z3 = 3xyz ........(1)

Since, (a + 2b + 3c) = 0,

⇒ (a)3 + (2b)3 + (3c)3 = 3 × a × 2b × 3c

⇒ a3 + 8b3 + 27c3 = 18abc.

Hence, proved that a3 + 8b3 + 27c3 = 18abc.

Question 6

If x+1x=3x + \dfrac{1}{x} = 3, find the value of (x3+1x3)\Big(x^3 + \dfrac{1}{x^3}\Big).

Answer

Given,

x+1x=3x + \dfrac{1}{x} = 3

Using identity,

⇒(x+1x)3=x3+1x3+3(x+1x)⇒33=x3+1x3+3×3⇒27=x3+1x3+9⇒x3+1x3=27−9=18.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 3^3 = x^3 + \dfrac{1}{x^3} + 3 \times 3 \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{x^3} + 9 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 27 - 9 = 18.

Hence, x3+1x3=18.x^3 + \dfrac{1}{x^3} = 18.

Question 7

If x−1x=5x - \dfrac{1}{x} = 5, find the value of (x3−1x3)\Big(x^3 - \dfrac{1}{x^3}\Big).

Answer

Given,

x−1x=5x - \dfrac{1}{x} = 5

Using identity,

⇒(x3−1x3)=(x−1x)3+3(x−1x)⇒(x3−1x3)=(5)3+3×5⇒(x3−1x3)=125+15⇒(x3−1x3)=140\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (5)^3 + 3 \times 5 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 125 + 15 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 140

Hence, x3−1x3=140.x^3 - \dfrac{1}{x^3} = 140.

Question 8

If x−2x=6x - \dfrac{2}{x} = 6, find the value of (x3−8x3)\Big(x^3 - \dfrac{8}{x^3}\Big).

Answer

Given,

⇒x−2x=6\Rightarrow x - \dfrac{2}{x} = 6

Upon cubing both sides we get :

⇒(x−2x)3=63⇒(x)3−(2x)3−3×x×2x×(x−2x)=216⇒(x)3−(2x)3−6×6=216⇒x3−8x3−36=216⇒x3−8x3=216+36⇒x3−8x3=252.\Rightarrow \Big(x - \dfrac{2}{x}\Big)^3 = 6^3 \\[1em] \Rightarrow (x)^3 - \Big(\dfrac{2}{x}\Big)^3 - 3 \times x \times \dfrac{2}{x} \times \Big(x - \dfrac{2}{x}\Big) = 216 \\[1em] \Rightarrow (x)^3 - \Big(\dfrac{2}{x}\Big)^3 - 6 \times 6 = 216 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} - 36 = 216 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} = 216 + 36 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} = 252.

Hence, x3−8x3=252.x^3 - \dfrac{8}{x^3} = 252.

Question 9

If x+1x=4x + \dfrac{1}{x} = 4, find the values of:

(i) (x3+1x3)\Big(x^3 + \dfrac{1}{x^3}\Big)

(ii) (x−1x)\Big(x - \dfrac{1}{x}\Big)

(iii) (x3−1x3)\Big(x^3 - \dfrac{1}{x^3}\Big)

Answer

(i) Given,

x+1x=4x + \dfrac{1}{x} = 4

Using identity,

⇒(x+1x)3=x3+1x3+3(x+1x)⇒43=x3+1x3+3×4⇒64=x3+1x3+12⇒x3+1x3=64−12=52.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 4^3 = x^3 + \dfrac{1}{x^3} + 3 \times 4 \\[1em] \Rightarrow 64 = x^3 + \dfrac{1}{x^3} + 12 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 64 - 12 = 52.

Hence, x3+1x3=52.x^3 + \dfrac{1}{x^3} = 52.

(ii) Given,

x+1x=4x + \dfrac{1}{x} = 4

Using identity,

⇒(x+1x)2−(x−1x)2=4⇒(4)2−(x−1x)2=4⇒16−(x−1x)2=4⇒(x−1x)2=16−4⇒(x−1x)2=12⇒(x−1x)=12⇒(x−1x)=±23.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (4)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 16 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 16 - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 12 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{12}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.

Hence, (x−1x)=±23.\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.

(iii) Given,

(x−1x)=±23\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}

Case 1:

(x−1x)=23\Big(x - \dfrac{1}{x}\Big) = 2\sqrt{3}

We know that,

⇒(x3−1x3)=(x−1x)3+3(x−1x)\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big)

Substituting values we get :

⇒(x3−1x3)=(23)3+3(23)⇒(x3−1x3)=(8×33)+(63)⇒(x3−1x3)=(243)+(63)⇒(x3−1x3)=303\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (2\sqrt{3})^3 + 3(2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (8 \times 3\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (24\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 30\sqrt{3} \\[1em]

Case 1:

(x−1x)=−23\Big(x - \dfrac{1}{x}\Big) = -2\sqrt{3}

We know that,

⇒(x3−1x3)=(x−1x)3+3(x−1x)\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big)

Substituting values we get :

⇒(x3−1x3)=(−23)3+3(−23)⇒(x3−1x3)=(−8×33)+(−63)⇒(x3−1x3)=(−243)−(63)⇒(x3−1x3)=−303\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-2\sqrt{3})^3 + 3(-2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-8 \times 3\sqrt{3}) + (-6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-24\sqrt{3}) - (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = -30\sqrt{3} \\[1em]

Hence, (x3−1x3)=±303.\Big(x^3 - \dfrac{1}{x^3}\Big) = \pm 30\sqrt{3}.

Question 10

If a2+1a2=23a^2 + \dfrac{1}{a^2} = 23, find the values of:

(i) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(ii) (a3+1a3)\Big(a^3 + \dfrac{1}{a^3}\Big)

Answer

(i) Given,

a2+1a2=23a^2 + \dfrac{1}{a^2} = 23

Using identity,

⇒(a+1a)2=a2+1a2+2⇒(a+1a)2=23+2⇒(a+1a)2=25⇒(a+1a)=25⇒(a+1a)=±5\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 23 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{25} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 5

Hence, (a+1a)=±5.\Big(a + \dfrac{1}{a}\Big) = \pm 5.

(ii) Given,

(a+1a)=±5.\Big(a + \dfrac{1}{a}\Big) = \pm 5.

Case 1:

(a+1a)=+5.\Big(a + \dfrac{1}{a}\Big) = +5.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒53=a3+1a3+3×5⇒125=a3+1a3+15⇒a3+1a3=125−15=110.\Rightarrow 5^3 = a^3 + \dfrac{1}{a^3} + 3 \times 5 \\[1em] \Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 125 - 15 = 110.

Case 2:

(a+1a)=−5.\Big(a + \dfrac{1}{a}\Big) = -5.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒(−5)3=a3+1a3+3×(−5)⇒−125=a3+1a3−15⇒a3+1a3=−125+15=−110.\Rightarrow (-5)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-5) \\[1em] \Rightarrow -125 = a^3 + \dfrac{1}{a^3} - 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -125 + 15 = -110.

Hence, a3+1a3=±110.a^3 + \dfrac{1}{a^3} = \pm 110.

Question 11

If a−1a=5a - \dfrac{1}{a} = \sqrt{5}, find the values of :

(i) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(ii) (a3+1a3)\Big(a^3 + \dfrac{1}{a^3}\Big)

Answer

(i) Given,

a−1a=5a - \dfrac{1}{a} = \sqrt{5}

Using identity,

⇒(a+1a)2−(a−1a)2=4⇒(a+1a)2−(5)2=4⇒(a+1a)2−5=4⇒(a+1a)2=4+5⇒(a+1a)2=9⇒(a+1a)=9⇒(a+1a)=±3\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (\sqrt{5})^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 5 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 5 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 9 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 3

Hence, (a+1a)=±3.\Big(a + \dfrac{1}{a}\Big) = \pm 3.

(ii) Given,

(a+1a)=±3.\Big(a + \dfrac{1}{a}\Big) = \pm 3.

Case 1:

(a+1a)=+3.\Big(a + \dfrac{1}{a}\Big) = +3.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒33=a3+1a3+3×3⇒27=a3+1a3+9⇒a3+1a3=27−9=18.\Rightarrow 3^3 = a^3 + \dfrac{1}{a^3} + 3 \times 3 \\[1em] \Rightarrow 27 = a^3 + \dfrac{1}{a^3} + 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 27 - 9 = 18.

Case 2:

(a+1a)=−3.\Big(a + \dfrac{1}{a}\Big) = -3.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒(−3)3=a3+1a3+3×(−3)⇒−27=a3+1a3−9⇒a3+1a3=−27+9=−18.\Rightarrow (-3)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-3) \\[1em] \Rightarrow -27 = a^3 + \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -27 + 9 = -18.

Hence, a3+1a3=±18.a^3 + \dfrac{1}{a^3} = \pm 18.

Question 12

If a2+1a2=27a^2 + \dfrac{1}{a^2} = 27, find the values of :

(i) (a−1a)\Big(a - \dfrac{1}{a}\Big)

(ii) (a3−1a3)\Big(a^3 - \dfrac{1}{a^3}\Big)

Answer

(i) Given,

a2+1a2=27a^2 + \dfrac{1}{a^2} = 27

Using identity,

⇒(a−1a)2=a2+1a2−2⇒(a−1a)2=27−2⇒(a−1a)2=25⇒(a−1a)2=25⇒(a−1a)2=±5\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 27 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \sqrt{25} \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \pm 5

Hence, (a−1a)2=±5.\Big(a - \dfrac{1}{a}\Big)^2 = \pm 5.

(ii) Given,

(a+1a)=±5.\Big(a + \dfrac{1}{a}\Big) = \pm 5.

Case 1:

(a+1a)=+5.\Big(a + \dfrac{1}{a}\Big) = +5.

We know that,

⇒(a3−1a3)=(a−1a)3+3×a×1a(a−1a)\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)

Substituting values we get :

⇒(a3−1a3)=(5)3+3(5)⇒(a3−1a3)=(125)+(15)⇒(a3−1a3)=140.\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (5)^3 + 3(5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (125) + (15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = 140.

Case 2:

(a+1a)=−5.\Big(a + \dfrac{1}{a}\Big) = -5.

We know that,

⇒(a3−1a3)=(a−1a)3+3×a×1a(a−1a)\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)

Substituting values we get :

⇒(a3−1a3)=(−5)3+3(−5)⇒(a3−1a3)=(−125)+(−15)⇒(a3−1a3)=−140.\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-5)^3 + 3(-5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-125) + (-15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = -140.

Hence, a3−1a3=±140a^3 - \dfrac{1}{a^3} = \pm 140

Question 13

If x2+125x2=835x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}, find the values of:

(i) (x+15x)\Big(x + \dfrac{1}{5x}\Big)

(ii) (x3+1125x3)\Big(x^3 + \dfrac{1}{125x^3}\Big)

Answer

(i) Given,

⇒x2+125x2=835⇒x2+125x2=435\Rightarrow x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5} \\[1em] \Rightarrow x^2 + \dfrac{1}{25x^2} = \dfrac{43}{5}

We know that,

⇒(x+15x)2=x2+(15x)2+2×x×15x⇒(x+15x)2=x2+125x2+25⇒(x+15x)2=435+25⇒(x+15x)2=455⇒(x+15x)2=9⇒(x+15x)=9⇒(x+15x)=±3\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \Big(\dfrac{1}{5x}\Big)^2 + 2 \times x \times \dfrac{1}{5x} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \dfrac{1}{25x^2} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{43}{5} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \pm 3 \\[1em]

Hence, (x+15x)=±3.\Big(x + \dfrac{1}{5x}\Big) = \pm 3.

(ii) We know that,

⇒(x+15x)3=(x)3+(15x)3+3×x×15x×(x+15x)⇒(x+15x)3=(x)3+(15x)3+35×(x+15x) .........(1)\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + 3 \times x \times \dfrac{1}{5x} \times \Big(x + \dfrac{1}{5x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + \dfrac{3}{5} \times \Big(x + \dfrac{1}{5x}\Big) \text{ .........(1)}

Given,

(x+15x)=±3.\Big(x + \dfrac{1}{5x}\Big) = \pm 3.

Case 1:

(x+15x)=+3.\Big(x + \dfrac{1}{5x}\Big) = +3.

Substituting values we get :

⇒(3)3=x3+1125x3+35×(3)⇒27=x3+1125x3+95⇒x3+1125x3=27−95⇒x3+1125x3=135−95⇒x3+1125x3=1265⇒x3+1125x3=2515\Rightarrow (3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(3) \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{125x^3} + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 27 - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{135-9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 25\dfrac{1}{5}

Case 2:

(x+15x)=−3.\Big(x + \dfrac{1}{5x}\Big) = -3.

Substituting values in equation (1), we get :

⇒(−3)3=x3+1125x3+35×(−3)⇒−27=x3+1125x3−95⇒x3+1125x3=−27+95⇒x3+1125x3=−135+95⇒x3+1125x3=−1265⇒x3+1125x3=−2515\Rightarrow (-3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(-3) \\[1em] \Rightarrow -27 = x^3 + \dfrac{1}{125x^3} - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -27 + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{-135 + 9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -\dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -25\dfrac{1}{5}

Hence, x3+125x3=±2515.x^3 + \dfrac{1}{25x^3} = \pm 25\dfrac{1}{5}.

Question 14

If (x+1x)2=3\Big(x + \dfrac{1}{x}\Big)^2 = 3, show that (x3+1x3)=0\Big(x^3 + \dfrac{1}{x^3}\Big) = 0.

Answer

Given,

⇒(x+1x)2=3⇒(x+1x)=±3.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{3}.

We know that,

⇒(x3+1x3)=(x+1x)3−3(x+1x)\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big)

Case 1 :

(x+1x)=+3\Big(x + \dfrac{1}{x}\Big) = + \sqrt{3}

Substituting values we get :

⇒(x3+1x3)=(3)3−3×3⇒(x3+1x3)=33−33=0.\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 3\sqrt{3} - 3\sqrt{3} = 0.

Case 2 :

(x+1x)=−3\Big(x + \dfrac{1}{x}\Big) = - \sqrt{3}

Substituting values we get :

⇒(x3+1x3)=(−3)3−3×−3⇒(x3+1x3)=−33+33=0.\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = -3\sqrt{3} + 3\sqrt{3} = 0.

Hence, proved that (x3+1x3)=0\Big(x^3 + \dfrac{1}{x^3}\Big) = 0.

Question 15

If ab=bc\dfrac{a}{b} = \dfrac{b}{c}, prove that (a + b + c)(a - b + c) = a2 + b2 + c2.

[Hint: Let ab=bc=k\dfrac{a}{b} = \dfrac{b}{c} = k, so b=ckb = ck and a=ck2a = ck^2.]

Answer

Given,

⇒ab=bc⇒a=bc×b⇒a×c=b2⇒ac=b2\Rightarrow \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow a = \dfrac{b}{c} \times b \\[1em] \Rightarrow a \times c = b^2 \\[1em] \Rightarrow ac = b^2

To prove,

(a + b + c)(a - b + c) = a2 + b2 + c2.

Solving L.H.S,

⇒ (a + b + c)(a - b + c)

⇒ a(a - b + c) + b(a - b + c) + c(a - b + c)

⇒ a2 - ab + ac + ab - b2 + bc + ca - bc + c2

= a2 + 2ac - b2 + c2

Substituting, ac = b2,

⇒ a2 + 2(b2) - b2 + c2

⇒ a2 + b2 + c2.

Since, L.H.S. = R.H.S.

Hence, proved that (a + b + c)(a - b + c) = a2 + b2 + c2.

Question 16

Find the product using suitable identities:

(i) (3a + 4b)(9a2 - 12ab + 16b2)

(ii) (y−6y)(y2+6+36y2)\Big(y - \dfrac{6}{y}\Big)\Big(y^2 + 6 + \dfrac{36}{y^2}\Big)

Answer

(i) Given,

⇒ (3a + 4b)(9a2 - 12ab + 16b2)

⇒ (3a + 4b)(3a)2 + (4b)2 - 3a × 4b

Using identity,

⇒ (a + b)(a2 - ab + b2) = (a3 - b3)

⇒ (3a)3 + (4b)3

Hence, (3a + 4b)(9a2 - 12ab + 16b2) = 27a3 + 64b3.

(ii) Using identity,

⇒ (a - b)(a2 + ab + b2) = (a3 - b3)

Given,

⇒(y−6y)(y2+6+36y2)⇒(y−6y)(y2+y×6y+(6y)2)⇒y3−(6y)3⇒y3−216y3.\Rightarrow \Big(y - \dfrac{6}{y}\Big)\Big(y^2 + 6 + \dfrac{36}{y^2}\Big) \\[1em] \Rightarrow \Big(y - \dfrac{6}{y}\Big)\Big(y^2 + y \times \dfrac{6}{y} + \Big(\dfrac{6}{y}\Big)^2\Big) \\[1em] \Rightarrow y^3 - \Big(\dfrac{6}{y}\Big)^3 \\[1em] \Rightarrow y^3 - \dfrac{216}{y^3}.

Hence, (y−6y)(y2+6+36y2)=y3−216y3\Big(y - \dfrac{6}{y}\Big) \Big(y^2 + 6 + \dfrac{36}{y^2}\Big) = y^3 - \dfrac{216}{y^3}.

Question 17

Simplify using suitable identity:

(3x - 5y - 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x)

Answer

Taking,

a = 3x, b = -5y and c = -4.

⇒ a2 + b2 + c2 - ab - bc - ca = (3x)2 + (-5y)2 + (-4)2 - 3x × (-5y) - (-5y) × (-4) - (-4) × 3x

⇒ a2 + b2 + c2 - ab - bc - ca = 9x2 + 25y2 + 16 + 15xy - 20y + 12x.

Using identity,

(a + b + c)(a2 + b2 + c2 - ab - bc - ca) = a3 + b3 + c3 - 3abc

⇒ (3x - 5y - 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x) = (3x)3 + (-5y)3 + (-4)3 - 3 × 3x × (-5y) × (-4)

⇒ (3x - 5y - 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x) = 27x3 - 125y3 - 64 - 180xy.

Hence, (3x - 5y - 4)(9x2 + 25y2 + 16 + 15xy - 20y + 12x) = 27x3 - 125y3 - 64 - 180xy.

Exercise 3C

Question 1

Write down the following products:

(x + 6)(x + 2)

Answer

(i) Given,

⇒ (x + 6)(x + 2)

⇒ (x2 + 2x + 6x + 12)

⇒ (x2 + 8x + 12).

Hence, (x + 6)(x + 2) = (x2 + 8x + 12).

Question 2

Write down the following products:

(x + 8)(x - 3)

Answer

Given,

⇒ (x + 8)(x - 3)

⇒ (x2 - 3x + 8x - 24)

⇒ (x2 + 5x - 24).

Hence, (x + 8)(x - 3) = (x2 + 5x - 24).

Question 3

Write down the following products:

(x - 5)(x - 7)

Answer

Given,

⇒ (x - 5)(x - 7)

⇒ (x2 - 7x - 5x + 35)

⇒ (x2 - 12x + 35).

Hence, (x - 5)(x - 7) = (x2 - 12x + 35).

Question 4

Write down the following products:

(2 - x)(4 - x)

Answer

Given,

⇒ (2 - x)(4 - x)

⇒ (8 - 2x - 4x + x2)

⇒ (8 - 6x + x2).

Hence, (2 - x)(4 - x) = (8 - 6x + x2).

Question 5

Write down the following products:

(y - 7)(y + 4)

Answer

Given,

⇒ (y - 7)(y + 4)

⇒ (y2 + 4y - 7y - 28)

⇒ (y2 - 3y - 28).

Hence, (y - 7)(y + 4) = (y2 - 3y - 28).

Question 6

Write down the following products:

(ab + 3)(ab - 2)

Answer

Given,

⇒ (ab + 3)(ab - 2)

⇒ (a2b2 - 2ab + 3ab - 6)

⇒ (a2b2 + ab - 6).

Hence, (ab + 3)(ab - 2) = (a2b2 + ab - 6).

Question 7

Write down the following products:

(5 - xy)(2 + xy)

Answer

Given,

⇒ (5 - xy)(2 + xy)

⇒ (10 + 5xy - 2xy - x2y2)

⇒ (10 + 3xy - x2y2).

Hence, (5 - xy)(2 + xy) = (10 + 3xy - x2y2).

Question 8

Write down the following products:

(x2 + 1)(x2 + 2)

Answer

Given,

⇒ (x2 + 1)(x2 + 2)

⇒ (x4 + 2x2 + x2 + 2)

⇒ (x4 + 3x2 + 2).

Hence, (x2 + 1)(x2 + 2) = (x4 + 3x2 + 2).

Question 9

Write down the following products:

(3 - x2)(5 + x2)

Answer

Given,

⇒ (3 - x2)(5 + x2)

⇒ (15 + 3x2 - 5x2 - x4)

⇒ (15 - 2x2 - x4).

Hence, (3 - x2)(5 + x2) = (15 - 2x2 - x4).

Question 10

Write down the following products:

(6 - x)(x + 5)

Answer

Given,

⇒ (6 - x)(x + 5)

⇒ (6x + 30 - x2 - 5x)

⇒ (30 - x2 + x).

Hence, (6 - x)(x + 5) = (30 - x2 + x).

Question 11

Write down the following products:

(2x + 3)(3x + 5)

Answer

Given,

⇒ (2x + 3)(3x + 5)

⇒ (6x2 + 10x + 9x + 15)

⇒ (6x2 + 19x + 15).

Hence, (2x + 3)(3x + 5) = (6x2 + 19x + 15).

Question 12

Write down the following products:

(7m - 2)(4m + 3)

Answer

Given,

⇒ (7m - 2)(4m + 3)

⇒ (28m2 + 21m - 8m - 6)

⇒ (28m2 + 13m - 6).

Hence, (7m - 2)(4m + 3) = (28m2 + 13m - 6).

Question 13

Write down the following products:

(2y - 3)(3y - 5)

Answer

Given,

⇒ (2y - 3)(3y - 5)

⇒ (6y2 - 10y - 9y + 15)

⇒ (6y2 - 19y + 15).

Hence, (2y - 3)(3y - 5) = (6y2 - 19y + 15).

Question 14

Write down the following products:

(3a2 - b2)(2a2 + 5b2)

Answer

Given,

⇒ (3a2 - b2)(2a2 + 5b2)

⇒ (6a4 + 15a2b2 - 2a2b2 - 5b4)

⇒ (6a4 + 13a2b2 - 5b4).

Hence, (3a2 - b2)(2a2 + 5b2) = 6a4 + 13a2b2 - 5b4.

Multiple Choice Questions

Question 1

If a + b = 12 and ab = 35, then (a − b)2 =

  1. 2

  2. 4

  3. 16

  4. 20

Answer

Given,

a + b = 12 and ab = 35

Using identity,

⇒ (a + b)2 - (a - b)2 = 4ab

⇒ (a − b)2 = (a + b)2 - 4ab

⇒ (a − b)2 = (12)2 - 4(35)

⇒ (a − b)2 = 144 - 140

⇒ (a − b)2 = 4

Hence, Option 2 is the correct option.

Question 2

If a + b = 16 and a − b = 2, then ab =

  1. 48

  2. 56

  3. 63

  4. 65

Answer

Given,

a + b = 16 and a - b = 2

Using identity,

⇒ (a + b)2 - (a - b)2 = 4ab

⇒ 162 - 22 = 4ab

⇒ 4ab = 256 - 4

⇒ 4ab = 252

⇒ ab = 2524\dfrac{252}{4}

⇒ ab = 63.

Hence, Option 3 is the correct option.

Question 3

If p+1p=52 then p2+1p2p + \dfrac{1}{p} = \dfrac{5}{2} \text{ then }p^2 + \dfrac{1}{p^2} =

  1. 254\dfrac{25}{4}

  2. 194\dfrac{19}{4}

  3. 94\dfrac{9}{4}

  4. 174\dfrac{17}{4}

Answer

Given,

p+1p=52p + \dfrac{1}{p} = \dfrac{5}{2}

Using identity,

⇒(p+1p)2=p2+1p2+2⇒(52)2=(p2+1p2)+2⇒254=(p2+1p2)+2⇒254−2=(p2+1p2)⇒(25−84)=(p2+1p2)⇒(p2+1p2)=174\Rightarrow \Big(p + \dfrac{1}{p}\Big)^2 = p^2 + \dfrac{1}{p^2} + 2 \\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^2 = \Big(p^2 + \dfrac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \dfrac{25}{4} = \Big(p^2 + \dfrac{1}{p^2}\Big) + 2 \\[1em] \Rightarrow \dfrac{25}{4} - 2 = \Big(p^2 + \dfrac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(\dfrac{25 - 8}{4}\Big) = \Big(p^2 + \dfrac{1}{p^2}\Big) \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2}\Big) = \dfrac{17}{4} \\[1em]

Hence, Option 4 is the correct option.

Question 4

If x+1x=5x + \dfrac{1}{x} = 5, then x−1xx - \dfrac{1}{x} =

  1. ±21\pm \sqrt{21}

  2. ±29\pm \sqrt{29}

  3. ±3\pm 3

  4. ±2\pm 2

Answer

Given,

x+1x=5x + \dfrac{1}{x} = 5

Using identity,

⇒(x+1x)2=x2+1x2+2×x×1x⇒(5)2=x2+1x2+2⇒25−2=x2+1x2⇒x2+1x2=23\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x}\\[1em] \Rightarrow (5)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow 25 - 2 = x^2 + \dfrac{1}{x^2} \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 23\\[1em]

We know that,

⇒(x−1x)2=x2+1x2−2⇒(x−1x)2=23−2⇒(x−1x)2=21⇒(x−1x)=±21\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 23 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 21 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm \sqrt{21} \\[1em]

Hence, Option 1 is the correct option.

Question 5

If y+1y=2y + \dfrac{1}{y} = 2, then y5+1y5y^5 + \dfrac{1}{y^5} =

  1. 5

  2. 3

  3. 2

  4. 1

Answer

Given,

y+1y=2y + \dfrac{1}{y} = 2

⇒y2+1y=2⇒y2+1=2y⇒y2−2y+1=0⇒(y−1)2=0⇒y−1=0⇒y=1\Rightarrow \dfrac{y^2 + 1}{y} = 2 \\[1em] \Rightarrow y^2 + 1 = 2y \\[1em] \Rightarrow y^2 - 2y + 1 = 0 \\[1em] \Rightarrow (y - 1)^2 = 0 \\[1em] \Rightarrow y-1 = 0 \\[1em] \Rightarrow y = 1

Substituting value of y, we get :

⇒y5+1y5=(1)5+1(1)5=1+1=2\Rightarrow y^5 + \dfrac{1}{y^5} = (1)^5 + \dfrac{1}{(1)^5} = 1 + 1 = 2.

Hence, option 3 is correct option.

Question 6

If a2−1a2=7a^2 - \dfrac{1}{a^2} = 7, then a4+1a4a^4 + \dfrac{1}{a^4} =

  1. 49

  2. 51

  3. 53

  4. 60

Answer

Given,

a2−1a2=7a^2 - \dfrac{1}{a^2} = 7

Upon squaring both sides,

⇒(a2−1a2)2=a4+1a4−2⇒(7)2=(a4+1a4)−2⇒49=(a4+1a4)−2⇒(a4+1a4)=49+2⇒(a4+1a4)=51\Rightarrow \Big(a^2 - \dfrac{1}{a^2}\Big)^2 = a^4 + \dfrac{1}{a^4} - 2 \\[1em] \Rightarrow (7)^2 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow 49 = \Big(a^4 + \dfrac{1}{a^4}\Big) - 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 49 + 2 \\[1em] \Rightarrow \Big(a^4 + \dfrac{1}{a^4}\Big) = 51 \\[1em]

Hence, Option 2 is the correct option.

Question 7

If x + y + z = 0, then x3 + y3 + z3 =

  1. xyz

  2. 3xyz

  3. 27xyz

  4. 0

Answer

Given,

x + y + z = 0

Using identity,

⇒ x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

⇒ x3 + y3 + z3 - 3xyz = (0) × (x2 + y2 + z2 - xy - yz - zx)

⇒ x3 + y3 + z3 - 3xyz = 0

⇒ x3 + y3 + z3 = 3xyz.

Hence, Option 2 is the correct option.

Question 8

If ab+ba=1\dfrac{a}{b} + \dfrac{b}{a} = 1 (a, b ≠ 0), then a3 + b3 =

  1. 0

  2. 1

  3. 2

  4. 8

Answer

Given,

⇒ab+ba=1⇒a2+b2ab=1⇒a2+b2=ab⇒a2+b2−ab=0.\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = 1 \\[1em] \Rightarrow \dfrac{a^2 + b^2}{ab} = 1 \\[1em] \Rightarrow a^2 + b^2 = ab \\[1em] \Rightarrow a^2 + b^2 - ab = 0.

We know that,

⇒ a3 + b3 = (a + b)(a2 - ab + b2)

⇒ a3 + b3 = (a + b)(0)

⇒ a3 + b3 = 0.

Hence, Option 1 is the correct option.

Question 9

If p+1p=x and p−1p=yp + \dfrac{1}{p} = x\text{ and }p - \dfrac{1}{p} = y, then the relation between x and y is :

  1. x2 = y2

  2. x2 + y2 = 4

  3. x2 - y2 = 4

  4. xy = 2

Answer

Given,

Upon squaring p+1p=xp + \dfrac{1}{p} = x we get,

⇒(p+1p)2=x2⇒(p2+1p2+2)=x2 ....(1)\Rightarrow \Big(p + \dfrac{1}{p}\Big)^2 = x^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) = x^2 \text{ ....(1)}

Upon squaring p−1p=yp - \dfrac{1}{p} = y we get,

⇒(p−1p)2=y2⇒(p2+1p2−2)=y2 ....(2)\Rightarrow \Big(p - \dfrac{1}{p}\Big)^2 = y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = y^2 \text{ ....(2)}

Subtracting (2) from (1) we get,

⇒(p2+1p2+2)−(p2+1p2−2)=x2−y2⇒(p2+1p2+2−p2−1p2+2)=x2−y2⇒x2−y2=4.\Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2\Big) - \Big(p^2 + \dfrac{1}{p^2} - 2\Big) = x^2 - y^2 \\[1em] \Rightarrow \Big(p^2 + \dfrac{1}{p^2} + 2 - p^2 - \dfrac{1}{p^2} + 2\Big) = x^2 - y^2 \\[1em] \Rightarrow x^2 - y^2 = 4.

Hence, Option 3 is the correct option.

Question 10

If a=1a−7a = \dfrac{1}{a - 7}, then a−1aa - \dfrac{1}{a} =

  1. 0

  2. 1

  3. 7

  4. 17\dfrac{1}{7}

Answer

Given,

⇒a=1a−7⇒a×(a−7)=1⇒a−7=1a⇒a−1a=7\Rightarrow a = \dfrac{1}{a - 7} \\[1em] \Rightarrow a \times (a-7) = 1 \\[1em] \Rightarrow a - 7 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 7

Hence, option 3 is correct option.

Question 11

2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51 =

  1. 1.44

  2. 0.44

  3. 1

  4. 0

Answer

Given,

⇒ 2.51 × 2.51 + 1.31 × 1.31 − 2.62 × 2.51

⇒ 2.512 + 1.312 - 2(1.31)(2.51)

⇒ 2.512 + 1.312 - 2(2.51)(1.31)

⇒ (2.51 - 1.31)2     [(a - b)2 = a2 + b2 - 2ab]

⇒ (1.2)2

⇒ 1.44

Hence, Option 1 is the correct option.

Question 12

If (x+1x)2=3\Big(x + \dfrac{1}{x}\Big)^2 = 3, then x3+1x3x^3 + \dfrac{1}{x^3} =

  1. 9

  2. 3

  3. 2

  4. 0

Answer

Given,

⇒(x+1x)2=3⇒(x+1x)=±3\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{3}

Case 1:

⇒(x+1x)=3\Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{3}

Using identity,

⇒(x3+1x3)=(x+1x)3−3(x+1x)⇒(x3+1x3)=(3)3−3(3)⇒(x3+1x3)=(33)−3(3)⇒(x3+1x3)=0\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (\sqrt{3})^3 - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (3\sqrt{3}) - 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0

Case 2:

⇒(x+1x)=−3\Rightarrow \Big(x + \dfrac{1}{x}\Big) = -\sqrt{3}

Using identity,

⇒(x3+1x3)=(x+1x)3−3(x+1x)⇒(x3+1x3)=(−3)3−3(−3)⇒(x3+1x3)=(−33)+3(3)⇒(x3+1x3)=0\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3(-\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-3\sqrt{3}) + 3(\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 0

Hence, Option 4 is the correct option.

Question 13

The value of ab if 3a + 5b = 15 and 9a2 + 25b2 = 75 is :

  1. 4

  2. 5

  3. 6

  4. 8

Answer

Given,

3a + 5b = 15 and 9a2 + 25b2 = 75

⇒ (3a + 5b)2 = (3a)2 + (5b)2 + 2 × (3a) × (5b)

⇒ (3a + 5b)2 = 9a2 + 25b2 + 30ab

⇒ 152 = 75 + 30ab

⇒ 225 = 75 + 30ab

⇒ 30ab = 225 - 75

⇒ 30ab = 150

⇒ ab = 15030\dfrac{150}{30}

⇒ ab = 5.

Hence, Option 2 is the correct option.

Question 14

If l + m − n = 9 and l2 + m2 + n2 = 31, then mn + nl − lm is :

  1. -25

  2. 25

  3. -2

  4. -5

Answer

Given,

l + m − n = 9

l2 + m2 + n2 = 31

Solving,

⇒ [(l + m) − (n)]2 = (l + m)2 + n2 - 2 × (l + m) × n

⇒ [(l + m) − (n)]2 = l2 + m2 + 2 × l × m + n2 - 2ln - 2mn

⇒ (9)2 = l2 + m2 + 2lm + n2 - 2ln - 2mn

⇒ 81 = l2 + m2 + n2 + 2lm - 2ln - 2mn

⇒ 81 = 31 + 2(lm - ln - mn)

⇒ 2(lm - ln - mn) = 81 - 31

⇒ (lm - ln - mn) = 502\dfrac{50}{2}

⇒ (lm - ln - mn) = 25

⇒ -(lm - ln - mn) = -25

⇒ (mn + nl - lm) = -25

Hence, Option 1 is the correct option.

Question 15

If a − b + c = 6 and a2 + b2 + c2 = 38, then ab + bc − ca is :

  1. 0

  2. 1

  3. -1

  4. not possible

Answer

Given,

a - b + c = 6

a2 + b2 + c2 = 38

We know that,

⇒ [(a - b) + (c)]2 = (a - b)2 + c2 + 2 × (a - b) × c

⇒ [(a - b) + (c)]2 = a2 + b2 - 2 × a × b + c2 + 2ac - 2bc

⇒ (6)2 = a2 + b2 + c2 - 2(ab - ac + bc)

⇒ 36 = 38 - 2(ab - ac + bc)

⇒ 2(ab + bc − ca) = 38 - 36

⇒ 2(ab + bc − ca) = 2

⇒ (ab + bc − ca) = 22\dfrac{2}{2}

⇒ (ab + bc − ca) = 1.

Hence, Option 2 is the correct option.

Assertion Reasoning Questions

Question 1

Assertion (A): (26)3 + (−15)3 + (−11)3 = 3 × 26 × 15 × 11.

Reason (R): If x + y + z = 0, then x3 + y3 + z3 = 3xyz

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

We know that,

⇒ x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)

If x + y + z = 0, then :

⇒ x3 + y3 + z3 - 3xyz = 0

⇒ x3 + y3 + z3 = 3xyz.

So, reason (R) is true.

⇒ 26 + (-15) + (-11)

⇒ 26 - 26

⇒ 0

Since, 26 + (-15) + (-11) = 0,

∴ (26)3 + (−15)3 + (−11)3 = 3 × 26 × -15 × -11

Assertion (A) is false.

Thus, A is false and R is true.

Hence, Option 2 is the correct option.

Question 2

Assertion (A): If x2−2x−1=0x^2 - 2x - 1 = 0, then x2+1x2=6x^2 + \dfrac{1}{x^2} = 6.
Reason (R): x2 - 2x - 1 can be written as (x - 1)2.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

Given,

⇒x2−2x−1=0⇒x2−1=2x⇒x2−1x=2⇒x2x−1x=2⇒x−1x=2\Rightarrow x^2 - 2x - 1 = 0 \\[1em] \Rightarrow x^2 - 1 = 2x \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 2 \\[1em] \Rightarrow \dfrac{x^2}{x} - \dfrac{1}{x} = 2 \\[1em] \Rightarrow x - \dfrac{1}{x} = 2 \\[1em]

Using identity,

(x−1x)2=x2+1x2−2\Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2

Substituting,

⇒(2)2=x2+1x2−2⇒4=x2+1x2−2⇒x2+1x2=4+2⇒x2+1x2=6.\Rightarrow (2)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 4 + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 6.

Assertion (A) is true.

⇒ (x - 1)2 = x2 + 12 - 2(x)(1)

⇒ (x - 1)2 = x2 - 2x + 1

Reason (R) is false.

A is true, R is false

Hence, Option 1 is the correct option.

Question 3

Assertion (A): (1 - 3x)3 can be expanded as 1 - 27x3 - 9x - 27x2.
Reason (R): (a - b)3 = a3 - b3 - 3ab(a - b)

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

Using identity,

(a - b)3 = a3 - b3 - 3ab(a - b)

So, reason (R) is true.

⇒ (1 - 3x)3 = 13 - (3x)3 - 3 × 1 × 3x (1 - 3x)

⇒ (1 - 3x)3 = 1 - 27x3 - 9x(1 - 3x)

⇒ (1 - 3x)3 = 1 - 27x3 - 9x + 27x2

So, assertion (A) is false.

Hence, Option 2 is the correct option.

Competency Focused Questions

Question 1

The value of (−28)3 + (18)3 + (10)3 is

  1. 15120

  2. -15120

  3. -5040

  4. none of these

Answer

Given,

(−28)3 + (18)3 + (10)3

We know that,

If a + b + c = 0, then a3 + b3 + c3 = 3abc.

Since, -28 + 18 + 10 = 0

⇒ (−28)3 + 183 + 103 = 3 × (-28) × 18 × 10

⇒ (−28)3 + 183 + 103 = -15120.

Hence, Option 2 is the correct option.

Question 2

If x + y = −4, the value of x3 + y3 − 12xy + 64 is :

  1. 1

  2. -1

  3. 4

  4. 0

Answer

Given,

x + y = -4

We know that,

⇒ x3 + y3 = (x + y)3 - 3xy(x + y)

⇒ x3 + y3 = (-4)3 - 3xy(-4)

⇒ x3 + y3 = -64 + 12xy

⇒ x3 + y3 - 12xy + 64 = 0.

Hence, Option 4 is the correct option.

Question 3

If a + b + c = 0, then (a+b)2ab+(b+c)2bc+(c+a)2ac\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac} is equal to :

  1. 0

  2. 1

  3. 3

  4. abc

Answer

Given,

a + b + c = 0

⇒ a + b = -c

⇒ b + c = -a

⇒ c + a = -b

Substituting the above values in (a+b)2ab+(b+c)2bc+(c+a)2ac\dfrac{(a + b)^2}{ab} + \dfrac{(b + c)^2}{bc} + \dfrac{(c + a)^2}{ac}, we get :

⇒(−c)2ab+(−a)2bc+(−b)2ca⇒c2ab+a2bc+b2ca⇒a3+b3+c3abc ......(1)\Rightarrow \dfrac{(-c)^2}{ab} + \dfrac{(-a)^2}{bc} + \dfrac{(-b)^2}{ca} \\[1em] \Rightarrow \dfrac{c^2}{ab} + \dfrac{a^2}{bc} + \dfrac{b^2}{ca} \\[1em] \Rightarrow \dfrac{a^3 + b^3 + c^3}{abc} \text{ ......(1)}

We know that,

If, a + b + c = 0 then a3 + b3 + c3 = 3abc

Substituting the value of a3 + b3 + c3 in (1), we get :

⇒3abcabc⇒3.\Rightarrow \dfrac{3abc}{abc} \\[1em] \Rightarrow 3.

Hence, option 3 is correct option.

Question 4

If ab+ba=−1\dfrac{a}{b} + \dfrac{b}{a} = -1 (a, b ≠ 0), then the value of a3 - b3 is :

  1. 12\dfrac{1}{2}

  2. 1

  3. -1

  4. 0

Answer

Given,

⇒ab+ba=−1⇒a2+b2ab=−1⇒a2+b2=−ab⇒a2+b2+ab=0\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = -1 \\[1em] \Rightarrow \dfrac{a^2 + b^2}{ab} = -1 \\[1em] \Rightarrow a^2 + b^2 = -ab \\[1em] \Rightarrow a^2 + b^2 + ab = 0

Using identity,

⇒ a3 - b3 = (a - b)(a2 + b2 + ab)

⇒ a3 - b3 = (a - b)(0)

⇒ a3 - b3 = 0.

Hence, option 4 is correct option.

Question 5

If x4+1x4=119x^4 + \dfrac{1}{x^4} = 119, x > 1, then find the value of x3−1x3x^3 - \dfrac{1}{x^3}

Answer

Given,

x4+1x4=119x^4 + \dfrac{1}{x^4} = 119, and x > 1.

x2+1x2x^2 + \dfrac{1}{x^2}

Using identity

(x2+1x2)2=x4+1x4+2\Big(x^2 + \dfrac{1}{x^2}\Big)^2 = x^4 + \dfrac{1}{x^4} + 2

Substitute the given value:

⇒(x2+1x2)2=119+2⇒(x2+1x2)2=121⇒x2+1x2=121⇒x2+1x2=±11\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 119 + 2 \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 121 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \sqrt{121} \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \pm 11

Since x>1,x2+1x2x \gt 1, x^2 + \dfrac{1}{x^2} must be positive.

So, x2+1x2=11x^2 + \dfrac{1}{x^2} = 11

x−1xx - \dfrac{1}{x}

Using identity

(x−1x)2=x2+1x2−2\Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2

Substitute the value:

⇒(x−1x)2=11−2⇒(x−1x)2=9⇒x−1x=±9⇒x−1x=±3\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 11 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm\sqrt{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 3

Since x>1,xx \gt 1, x is greater than 1x\dfrac{1}{x}, so x−1xx - \dfrac{1}{x} must be positive.

So, x−1x=3x - \dfrac{1}{x} = 3

x3−1x3x^3 - \dfrac{1}{x^3}

By using the identity:

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Let a=xa = x and b=1xb = \dfrac{1}{x}.

⇒x3−1x3=(x−1x)(x2+x⋅1x+1x2)⇒x3−1x3=(x−1x)(x2+1x2+1)⇒x3−1x3=(3)(11+1)⇒x3−1x3=(3)(12)⇒x3−1x3=36\Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + x \cdot \dfrac{1}{x} + \dfrac{1}{x^2}\Big) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + \dfrac{1}{x^2} + 1\Big)\\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(11 + 1) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(12) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 36

Hence, x3−1x3=36x^3 - \dfrac{1}{x^3} = 36.

Question 6

If 1x+y=1x+1y\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y} (x ≠ 0, y ≠ 0, x ≠ y), then find the value of x3 - y3

Answer

Given,

1x+y=1x+1y\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y}

⇒1x+y=y+xxy⇒xy=(x+y)(x+y)⇒xy=x2+y2+2xy⇒0=x2+y2+2xy−xy⇒x2+xy+y2=0\Rightarrow \dfrac{1}{x + y} = \dfrac{y + x}{xy} \\[1em] \Rightarrow xy = (x + y)(x + y) \\[1em] \Rightarrow xy = x^2 + y^2 + 2xy \\[1em] \Rightarrow 0 = x^2 + y^2 + 2xy - xy \\[1em] \Rightarrow x^2 + xy + y^2 = 0

Using identity,

x3 - y3 = (x - y)(x2 + xy +y2)

x3 - y3 = (x - y)(0)

x3 - y3 = 0

Hence, x3 - y3 = 0.

Question 7

If x2+y2+1x2+1y2=4x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} = 4, then find the value of x2 + y2

Answer

Given,

⇒x2+y2+1x2+1y2=4⇒x2+y2+1x2+1y2−4=0⇒(x2+1x2−2)+(y2+1y2−2)=0\Rightarrow x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} = 4 \\[1em] \Rightarrow x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} - 4 = 0 \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} - 2\Big) + \Big(y^2 + \dfrac{1}{y^2} - 2\Big) = 0

Using identity,

a2+1a2−2=(a−1a)2a^2 + \dfrac{1}{a^2} - 2 = \Big(a - \dfrac{1}{a}\Big)^2.

(x−1x)2+(y−1y)2=0\Big(x - \dfrac{1}{x}\Big)^2 + \Big(y - \dfrac{1}{y}\Big)^2 = 0

Hence,

(x−1x)2=0 and (y−1y)2=0\Big(x - \dfrac{1}{x}\Big)^2 = 0 \text{ and } \Big(y - \dfrac{1}{y}\Big)^2 = 0

x−1x=0 and y−1y=0x - \dfrac{1}{x} = 0 \text{ and } y - \dfrac{1}{y} = 0

Solving for x: ⇒x=1x⇒x2=1\Rightarrow x = \dfrac{1}{x} \\[1em] \Rightarrow x^2 = 1

Solving for y: ⇒y=1y⇒y2=1\Rightarrow y = \dfrac{1}{y} \\[1em] \Rightarrow y^2 = 1

Solving for x2 + y2:

x2 + y2 = 1 + 1

x2 + y2 = 2

Hence, x2 + y2 = 2.

PrevNext