KnowledgeBoat Logo
|
OPEN IN APP

Chapter 3

Expansions

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 3.1

Question 1(i)

By using standard formulae, expand the following:

(2x + 7y)2

Answer

(2x + 7y)2 = (2x)2 + 2(2x)(7y) + (7y)2

= 4x2 + 28xy + 49y2

Question 1(ii)

By using standard formulae, expand the following:

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

Answer

(12x+23y)2=(12x)2+2(12x)(23y)+(23y)2=14x2+23xy+49y2\Big(\dfrac{1}{2}x + \dfrac{2}{3}y\Big)^2 = \Big(\dfrac{1}{2}x\Big)^2 + 2\Big(\dfrac{1}{2}x\Big)\Big(\dfrac{2}{3}y\Big) + \Big(\dfrac{2}{3}y\Big)^2 \\[1em] = \dfrac{1}{4}x^2 + \dfrac{2}{3}xy + \dfrac{4}{9}y^2

Question 2(i)

By using standard formulae, expand the following:

(3x+12x)2\Big(3x + \dfrac{1}{2x} \Big)^2

Answer

(3x+12x)2=(3x)2+2(3x)(12x)+(12x)2=9x2+3+(14x2)\Big(3x + \dfrac{1}{2x} \Big)^2 = (3x)^2 + 2 (3x)\Big(\dfrac{1}{2x}\Big) + \Big(\dfrac{1}{2x}\Big)^2 \\[1em] = 9x^2 + 3 + \Big(\dfrac{1}{4x^2}\Big)

Question 2(ii)

By using standard formulae, expand the following:

(3x2y + 5z)2

Answer

(3x2y + 5z)2 = (3x2y)2 + 2 (3x2y)(5z) + (5z)2

= 9x4y2 + 30x2yz + 25z2

Question 3(i)

By using standard formulae, expand the following:

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

Answer

(3x−12x)2=(3x)2−2(3x)(12x)+(12x)2=9x2−3+(14x2)\Big(3x - \dfrac{1}{2x} \Big)^2 = (3x)^2 - 2 (3x)\Big(\dfrac{1}{2x}\Big) + \Big(\dfrac{1}{2x}\Big)^2 \\[1em] = 9x^2 - 3 + \Big(\dfrac{1}{4x^2}\Big)

Question 3(ii)

By using standard formulae, expand the following:

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

Answer

(12x−32y)2=(12x)2−2(12x)(32y)+(32y)2=14x2−32xy+94y2\Big(\dfrac{1}{2}x - \dfrac{3}{2}y\Big)^2 = \Big(\dfrac{1}{2}x\Big)^2 - 2\Big(\dfrac{1}{2}x\Big)\Big(\dfrac{3}{2}y\Big) + \Big(\dfrac{3}{2}y\Big)^2 \\[1em] = \dfrac{1}{4}x^2 - \dfrac{3}{2}xy + \dfrac{9}{4}y^2

Question 4(i)

By using standard formulae, expand the following:

(x+3)(x+5)

Answer

We know that (x+a)(x+b) = x2 + (a + b)x + ab

∴ (x+3)(x+5) = x2 + (3 + 5)x + (3)(5) = x2 + 8x + 15

Question 4(ii)

By using standard formulae, expand the following:

(x+3)(x-5)

Answer

We know that (x+a)(x-b) = x2 + (a - b)x - ab

∴ (x+3)(x-5) = x2 + (3 - 5)x - (3)(5) = x2 - 2x - 15

Question 4(iii)

By using standard formulae, expand the following:

(x-7)(x+9)

Answer

We know that (x-a)(x+b) = x2 - (a - b)x - ab

∴ (x-7)(x+9) = x2 - (7 - 9)x - (7)(9) = x2 + 2x - 63

Question 4(iv)

By using standard formulae, expand the following:

(x-2y)(x-3y)

Answer

We know that (x-a)(x-b) = x2 - (a + b)x + ab

∴ (x-2y)(x-3y) = x2 - (2y + 3y)x + (2y)(3y)

= x2 - 5xy + 6y2

Question 5(i)

By using standard formulae, expand the following:

(x - 2y - z)2

Answer

(x - 2y - z)2 = [(x) + (-2y) + (-z)]2

= (x)2 + (-2y)2 + (-z)2 + 2 [(x)(-2y) + (-2y)(-z) + (-z)(x)]

= x2 + 4y2 + z2 + 2[-2xy + 2yz - xz]

= x2 + 4y2 + z2 - 4xy + 4yz - 2xz

Question 5(ii)

By using standard formulae, expand the following:

(2x - 3y + 4z)2

Answer

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

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

= 4x2 + 9y2 + 16z2 + 2[-6xy - 12yz + 8xz]

= 4x2 + 9y2 + 16z2 - 12xy - 24yz + 16xz

Question 6(i)

By using standard formulae, expand the following:

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

Answer

(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\Big(2x + \dfrac{3}{x} - 1\Big)^2 = \Big[\Big(2x\Big) + \Big(\dfrac{3}{x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = (2x)^2 + \Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 \Big[\Big(2x\Big)\Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(2x\Big) \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 2\Big[6 - \dfrac{3}{x} - 2x \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em]

Question 6(ii)

By using standard formulae, expand the following:

(23x−32x−1)2\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2

Answer

(23x−32x−1)2=[23x+(−32x)+(−1)]2=(23x)2+(−32x)2+(−1)2+2[(23x)(−32x)+(−32x)(−1)+(−1)(23x)]=49x2+94x2+1+2[−1+32x−23x]=49x2+94x2+1−2+3x−43x=49x2+94x2−1+3x−43x\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2 = \Big[\dfrac{2}{3}x + \Big(-\dfrac{3}{2x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = \Big(\dfrac{2}{3}x\Big)^2 + \Big(-\dfrac{3}{2x}\Big)^2 + (-1)^2 + 2 \Big[\Big(\dfrac{2}{3}x\Big)\Big(-\dfrac{3}{2x}\Big) + \Big(-\dfrac{3}{2x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(\dfrac{2}{3}x\Big) \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 + 2\Big[-1 + \dfrac{3}{2x} - \dfrac{2}{3}x \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 - 2 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} -1 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em]

Question 7(i)

By using standard formulae, expand the following:

(x+2)3

Answer

(x+2)3 = x3 + (2)3 + 3(x)(2)(x+2)

= x3 + 8 + 6x2 + 12x

Question 7(ii)

By using standard formulae, expand the following:

(2a+b)3

Answer

(2a+b)3 = (2a)3 + (b)3 + 3(2a)(b)(2a+b)

= 8a3 + b3 + 12a2b + 6ab2

Question 8(i)

By using standard formulae, expand the following:

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

Answer

(3x+1x)3=(3x)3+(1x)3+3(3x)(1x)(3x+1x)=27x3+1x3+27x+9x\Big(3x + \dfrac{1}{x}\Big)^3 = (3x)^3 + \Big(\dfrac{1}{x}\Big)^3 + 3(3x)\Big(\dfrac{1}{x}\Big)\Big(3x + \dfrac{1}{x}\Big) \\[1em] = 27x^3 + \dfrac{1}{x^3} + 27x + \dfrac{9}{x}

Question 8(ii)

By using standard formulae, expand the following:

(2x-1)3

Answer

(2x-1)3 = (2x)3 - (1)3 - 3(2x)(1)(2x-1)

= 8x3 - 1 -12x2 + 6x

Question 9(i)

By using standard formulae, expand the following:

(5x-3y)3

Answer

(5x-3y)3 = (5x)3 - (3y)3 - 3(5x)(3y)(5x-3y)

= 125x3 - 27y3 - 225x2y + 135xy2

Question 9(ii)

By using standard formulae, expand the following:

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

Answer

(2x−13y)3=(2x)3−(13y)3−3(2x)(13y)(2x−13y)=8x3−127y3−4x2y+2x3y2\Big(2x - \dfrac{1}{3y}\Big)^3 = (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3 - 3(2x)\Big(\dfrac{1}{3y}\Big)\Big(2x - \dfrac{1}{3y}\Big) \\[1em] = 8x^3 - \dfrac{1}{27y^3} - \dfrac{4x^2}{y} + \dfrac{2x}{3y^2}

Question 10(i)

Simplify the following:

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

Answer

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

Question 10(ii)

Simplify the following:

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

Answer

(a+1a)2−(a−1a)2=[(a)2+2(a)(1a)+(1a)2]−[(a)2−2(a)(1a)+(1a)2]=[a2+2+1a2]−[a2−2+1a2]=a2+2+1a2−a2+2−1a2=4\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = \Big[\Big(a\Big)^2 + 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2 \Big] - \Big[\Big(a\Big)^2 - 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2\Big] \\[1em] = \Big[a^2 + 2 + \dfrac{1}{a^2} \Big] - \Big[a^2 - 2 + \dfrac{1}{a^2} \Big] \\[1em] = a^2 + 2 + \dfrac{1}{a^2} - a^2 + 2 - \dfrac{1}{a^2} = 4

Question 11(i)

Simplify the following:

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

Answer

(3x-1)2 - (3x-2)(3x+1) = [(3x-1)2] - [(3x-2)(3x+1)]

= [(3x)2 - 2(3x)(1) + (1)2] - [(3x)2 + 3x - 6x - 2]

= (9x2 - 6x + 1) - (9x2 - 3x - 2)

= 9x2 - 6x + 1 - 9x2 + 3x + 2

= -3x + 3

= 3 - 3x

Question 11(ii)

Simplify the following:

(4x+3y)2 - (4x-3y)2 - 48xy

Answer

(4x+3y)2 - (4x-3y)2 - 48xy = [(4x)2 + 2(4x)(3y) + (3y)2] - [(4x)2 - 2(4x)(3y) + (3y)2] - 48xy

= [16x2 + 24xy + 9y2] - [16x2 - 24xy + 9y2] - 48xy

= 16x2 + 24xy + 9y2 - 16x2 + 24xy - 9y2 - 48xy

= 48xy - 48xy = 0

Question 12(i)

Simplify the following:

(7p+9q)(7p-9q)

Answer

(7p+9q)(7p-9q) = (7p)2 - (9q)2

= 49p2 - 81q2

Question 12(ii)

Simplify the following:

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

Answer

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

Question 13(i)

Simplify the following:

(2x - y + 3)(2x - y - 3)

Answer

(2x - y + 3)(2x - y - 3)

Let (2x - y) = a

Then, the given expression = (a+3)(a-3) = (a)2 - (3)2

= a2 - 9 = (2x-y)2 - 9 = (2x)2 - 2(2x)(y) + y2 - 9

= 4x2 - 4xy + y2 - 9

Question 13(ii)

Simplify the following:

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

Answer

(3x+y-5)(3x-y-5) = (3x-5+y)(3x-5-y)

Let (3x-5)=a

Then, the given expression = (a+y)(a-y)

= a2 - y2 = (3x-5)2 - y2

= [(3x)2 - 2(3x)(5) + (5)2] - y2

= 9x2 - 30x + 25 - y2

Question 14(i)

Simplify the following:

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

Answer

(x+2x−3)(x−2x−3)\Big(x + \dfrac{2}{x} - 3\Big)\Big(x - \dfrac{2}{x} - 3\Big) \\[1em]

Let (x - 3) = a

Then, the given expression = (a+2x)(a−2x)=(a)2−(2x)2=a2−4x2=(x−3)2−4x2=x2−2(x)(3)+32−4x2x2−6x+9−4x2\Big(a+\dfrac{2}{x}\Big)\Big(a-\dfrac{2}{x}\Big) = \Big(a\Big)^2 - \Big(\dfrac{2}{x}\Big)^2 \\[1em] = a^2 - \dfrac{4}{x^2} = (x-3)^2 - \dfrac{4}{x^2} = x^2 -2(x)(3) + 3^2 - \dfrac{4}{x^2} \\[1em] x^2 - 6x + 9 - \dfrac{4}{x^2}

Question 14(ii)

Simplify the following:

(5-2x) (5+2x) (25+4x2)

Answer

(5-2x)(5+2x)(25+4x2) = [(5)2 - (2x)2] (25+4x2)

= (25-4x2)(25+4x2)

= (25)2 - (4x2)2 = 625 - 16x4

Question 15(i)

Simplify the following:

(x+2y+3)(x+2y+7)

Answer

(x+2y+3)(x+2y+7)

Let (x+2y)=z

Then the given expression = (z+3)(z+7)

We know that (x+a)(x+b) = x2 + (a + b)x + ab

∴ (z+3)(z+7) = z2 + (3+7)z + (3)(7)

= z2 + 10z + 21

= (x+2y)2 + 10(x+2y) + 21

= x2 + 2(x)(2y) + (2y)2 + 10x + 20y + 21

= x2 + 4xy + 4y2 + 10x + 20y + 21

= x2 + 10x + 4y2 + 20y + 4xy + 21

Question 15(ii)

Simplify the following:

(2x+y+5)(2x+y-9)

Answer

(2x+y+5)(2x+y-9)

Let (2x+y)=z

Then the given expression = (z+5)(z-9)

We know that (x+a)(x-b) = x2 + (a - b)x - ab

∴ (z+5)(z-9) = z2 + (5-9)z - (5)(9)

= z2 - 4z - 45

= (2x+y)2 - 4(2x+y) - 45

= (2x)2 + 2(2x)(y) + (y)2 - 8x - 4y - 45

= 4x2 + 4xy + y2 - 8x - 4y - 45

= 4x2 - 8x + y2 - 4y + 4xy - 45

Question 15(iii)

Simplify the following:

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

Answer

Let (x - 2y) = z

Then the given expression = (z - 5)(z + 3)

We know that (x - a)(x + b) = x2 - (a - b)x - ab

∴ (z - 5)(z + 3) = z2 - (5 - 3)z - (5)(3)

= z2 - 2z - 15

= (x - 2y)2 - 2(x - 2y) - 15

= (x)2 - 2(x)(2y) + (2y)2 - 2x + 4y - 15

= x2 - 4xy + 4y2 - 2x + 4y - 15

Question 15(iv)

Simplify the following:

(3x - 4y - 2)(3x - 4y - 6)

Answer

Let (3x - 4y) = z

Then the given expression = (z - 2)(z - 6)

We know that (x - a)(x - b) = x2 - (a + b)x + ab

∴ (z - 2)(z - 6) = z2 - (2 + 6)z + (2)(6)

= z2 - 8z + 12

= (3x - 4y)2 - 8(3x - 4y) + 12

= (3x)2 - 2(3x)(4y) + (4y)2 - 24x + 32y + 12

= 9x2 - 24xy + 16y2 - 24x + 32y + 12

Question 16(i)

Simplify the following:

(2p + 3q)(4p2 - 6pq + 9q2)

Answer

(2p + 3q)(4p2 - 6pq + 9q2) = (2p + 3q)[(2p)2 - (2p)(3q) + (3q)2]

We know that (a + b)(a2 - ab + b2) = a3 + b3

∴ (2p + 3q)[(2p)2 - (2p)(3q) + (3q)2] = (2p)3 + (3q)3

= 8p3 + 27q3

Question 16(ii)

Simplify the following:

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

Answer

We know that (a + b)(a2 - ab + b2) = a3 + b3

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

=x3+1x3= x^3 + \dfrac{1}{x^3}

Question 17(i)

Simplify the following:

(3p - 4q)(9p2 + 12pq + 16q2)

Answer

(3p - 4q)(9p2 +12pq + 16q2) = (3p - 4q)[(3p)2 + (3p)(4q) + (4q)2]

We know that (a - b)(a2 + ab + b2) = a3 - b3

∴ (3p - 4q)[(3p)2 + (3p)(4q) + (4q)2] = (3p)3 - (4q)3

= 27p3 - 64q3

Question 17(ii)

Simplify the following:

(x−3x)(x2+3+9x2)\Big(x - \dfrac{3}{x}\Big)\Big(x^2 + 3 + \dfrac{9}{x^2}\Big)

Answer

We know that (a - b)(a2 + ab + b2) = a3 - b3

∴ Given Expression = (x−3x)(x2+3+9x2)\Big(x - \dfrac{3}{x}\Big)\Big(x^2 + 3 + \dfrac{9}{x^2}\Big)

=x3−(3x)3=x3−27x3= x^3 - \Big(\dfrac{3}{x}\Big)^3 \\[1em] = x^3 - \dfrac{27}{x^3}

Question 18

Simplify the following:

(2x + 3y + 4z)(4x2 + 9y2 + 16z2 - 6xy - 12yz - 8zx)

Answer

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

∴ The Given Expression = (2x + 3y + 4z)[(2x)2 + (3y)2 + (4z)2 - (2x)(3y) - (3y)(4z) - (4z)(2x)]

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

= 8x3 + 27y3 + 64z3 - 72xyz

Question 19(i)

Find the product of the following:

(x + 1)(x + 2)(x + 3)

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

∴ (x + 1)(x + 2)(x + 3) = x3 + (1 + 2 + 3)x2 + [(1)(2) + (2)(3) + (3)(1)]x + (1)(2)(3)

= x3 + 6x2 + (2 + 6 + 3)x + 6

= x3 + 6x2 + 11x + 6

Question 19(ii)

Find the product of the following:

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

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

∴ (x - 2)(x - 3)(x + 4) = x3 + [(-2) + (-3) + 4]x2 + [(-2)(-3) + (-3)(4) + (4)(-2)]x + (-2)(-3)(4)

= x3 - x2 + (6 -12 - 8)x + 24

= x3 - x2 - 14x + 24

Question 20

Find the coefficient of x2 and x in the product of (x - 3)(x + 7)(x - 4)

Answer

We know that (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

Comparing (x - 3)(x + 7)(x - 4) with (x + a)(x + b)(x + c).
Here a = -3, b = 7, c = -4.

∴ Coefficient of x2 = a + b + c = (-3) + 7 + (-4) = 0 and
    coefficient of x = ab + bc + ca

    = (-3) x 7 + 7 x (-4) + (-4) x (-3)

    = -21 - 28 + 12 = -37

Question 21

If a2 + 4a + x = (a + 2)2, find the value of x

Answer

Given,
    a2 + 4a + x = (a + 2)2

⇒ a2 + 4a + x = a2 + 2(a)(2) + 22

⇒ a2 - a2 +4a - 4a + x = 4

⇒ x = 4

Question 22(i)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(101)2

Answer

(101)2 = (100 + 1)2

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

(100 + 1)2 = (100)2 + 2(100)(1) + (1)2

= 10000 + 200 + 1 = 10201

Question 22(ii)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(1003)2

Answer

(1003)2 = (1000 + 3)2

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

(1000 + 3)2 = (1000)2 + 2(1000)(3) + (3)2

= 1000000 + 6000 + 9 = 1006009

Question 22(iii)

Use (a+b)2 = a2 + 2ab + b2 to evaluate the following:

(10.2)2

Answer

(10.2)2 = (10 + 0.2)2

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

(10 + 0.2)2 = (10)2 + 2(10)(0.2) + (0.2)2

= 100 + 4 + 0.04 = 104.04

Question 23(i)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(99)2

Answer

(99)2 = (100 - 1)2

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

(100 - 1)2 = (100)2 - 2(100)(1) + (1)2

= 10000 - 200 + 1 = 9801

Question 23(ii)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(997)2

Answer

(997)2 = (1000 - 3)2

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

(1000 - 3)2 = (1000)2 - 2(1000)(3) + (3)2

= 1000000 - 6000 + 9 = 994009

Question 23(iii)

Use (a - b)2 = a2 - 2ab + b2 to evaluate the following:

(9.8)2

Answer

(9.8)2 = (10 - 0.2)2

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

(10 - 0.2)2 = (10)2 - 2(10)(0.2) + (0.2)2

= 100 - 4 + 0.04 = 96.04

Question 24(i)

By using suitable identities, evaluate the following:

(103)3

Answer

(103)3 = (100 + 3)3

Using (a + b)3 = a3 + b3 + 3(a)(b)(a + b),

(100 + 3)3 = (100)3 + (3)3 + 3(100)(3)(100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727

Question 24(ii)

By using suitable identities, evaluate the following:

(99)3

Answer

(99)3 = (100 - 1)3

Using (a - b)3 = a3 - b3 - 3(a)(b)(a - b),

(100 - 1)3 = (100)3 - (1)3 - 3(100)(1)(100 - 1)

= 1000000 - 1 - 300(99)

= 1000000 - 1 - 29700

= 970299

Question 24(iii)

By using suitable identities, evaluate the following:

(10.1)3

Answer

(10.1)3 = (10 + 0.1)3

Using (a + b)3 = a3 + b3 + 3(a)(b)(a + b),

(10 + 0.1)3 = (10)3 + (0.1)3 + 3(10)(0.1)(10.1)

= 1000 + 0.001 + 30.3

= 1030.301

Question 25

If 2a - b + c = 0, prove 4a2 - b2 + c2 + 4ac = 0

Answer

Given,
     2a - b + c = 0
⇒ (2a + c) = b

On squaring both sides,

    (2a + c)2 = b2

⇒ 4a2 + c2 + 2(2a)(c) = b2

⇒ 4a2 - b2 + c2 + 4ac = 0

Hence proved.

Question 26

If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc

Answer

We know that if a + b + c = 0 then a3 + b3 + c3 = 3abc

Given,

a + b + 2c = 0

∴ (a)3 + (b)3 + (2c)3 = 3(a)(b)(2c)

⇒ a3 + b3 + 8c3 = 6abc

Hence Proved.

Question 27

If x + 2y - 3 = 0, then find the value of x3 + 8y3 + 6x2y + 12xy2 - 125.

Answer

Given,

x + 2y - 3 = 0

⇒ x + 2y = 3

Simplifying,

⇒ x3 + 8y3 + 6x2y + 12xy2 - 125

⇒ x3 + (2y)3 + 6xy(x + 2y) - 125

⇒ x3 + (2y)3 + 3.x.2y.(x + 2y) - 125

⇒ (x + 2y)3 - 125

⇒ 33 - 125

⇒ 27 - 125

⇒ -98.

Hence, the value of x3 + 8y3 + 6x2y + 12xy2 - 125 = -98.

Question 28

If a + b + c = 0, then find the value of a2bc+b2ca+c2ab\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}.

Answer

We know if a + b + c = 0, a3 + b3 + c3 = 3abc.

On dividing each side by abc,
⇒a3abc+b3abc+c3abc=3abcabc⇒a2bc+b2ca+c2ab=3\phantom{\Rightarrow} \dfrac{a^3}{abc} + \dfrac{b^3}{abc} + \dfrac{c^3}{abc} = \dfrac{3abc}{abc} \\[1em] \Rightarrow \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = 3

∴ Value of a2bc+b2ca+c2ab=3\bold{\dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab} = 3}

Question 29

If x + y = 4, find the value of x3 + y3 + 12xy - 64

Answer

Using (x + y)3 = x3 + y3 + 3(x)(y)(x + y)

Putting x + y = 4 in above:

    x3 + y3 + 3xy(4) = (4)3

⇒ x3 + y3 + 12xy = 64

⇒ x3 + y3 + 12xy - 64 = 0

∴ Value of x3 + y3 + 12xy - 64 = 0

Question 30

Without actually calculating the cubes, find the values of:

(i) (27)3 + (-17)3 + (-10)3

(ii) (-28)3 + (15)3 + (13)3

Answer

(i) Let a = 27, b = -17, c = -10

Then,

a + b + c = 37 - 17 - 10 = 0

Thus, a3 + b3 + c3 = 3abc

∴ (27)3 + (-17)3 + (-10)3 = 3(27)(-17)(-10) = 13770

(ii) Let a = -28, b = 15, c = 13

Then,

a + b + c = -28 + 15 + 13 = 0

Thus, a3 + b3 + c3 = 3abc

∴ (-28)3 + (15)3 + (13)3 = 3(-28)(15)(13) = -16380

Question 31

Using suitable identity, find the value of :

86×86×86+14×14×1486×86−86×14+14×14\dfrac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14}

Answer

Let x = 86 and y = 14.

Hence, above equation can be written as,

⇒x3+y3x2−xy+y2=(x+y)(x2−xy+y2)(x2−xy+y2)=x+y=86+14=100.\Rightarrow \dfrac{x^3 + y^3}{x^2 - xy + y^2} \\[1em] = \dfrac{(x + y)(x^2 - xy + y^2)}{(x^2 - xy + y^2)} \\[1em] = x + y \\[1em] = 86 + 14 \\[1em] = 100.

Hence, value of 86×86×86+14×14×1486×86−86×14+14×14\dfrac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14} = 100.

Exercise 3.2

Question 1

If x - y = 8 and xy = 5, find x2 + y2.

Answer

We know that,

(x - y)2 = x2 - 2xy + y2

⇒ x2 + y2 = (x - y)2 + 2xy.

Substituting values we get,

⇒ x2 + y2 = (8)2 + 2 × 5
⇒ x2 + y2 = 64 + 10
⇒ x2 + y2 = 74.

Hence, x2 + y2 = 74.

Question 2

If x + y = 10 and xy = 21, find 2(x2 + y2).

Answer

We know that,

(x + y)2 = x2 + 2xy + y2

⇒ x2 + y2 = (x + y)2 - 2xy.

⇒ 2(x2 + y2) = 2[(x + y)2 - 2xy]

Substituting values we get,

⇒ 2(x2 + y2) = 2[(10)2 - 2 × 21]

⇒ 2(x2 + y2) = 2(100 - 42)

⇒ 2(x2 + y2) = 2 × 58

⇒ 2(x2 + y2) = 116.

Hence, 2(x2 + y2) = 116.

Question 3

If 2a + 3b = 7 and ab = 2, find 4a2 + 9b2.

Answer

We know that,

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

∴ 4a2 + 9b2 = (2a)2 + (3b)2 = (2a + 3b)2 - 12ab.

Substituting values we get,

⇒ 4a2 + 9b2 = (7)2 - 12 × 2

⇒ 4a2 + 9b2 = 49 - 24

⇒ 4a2 + 9b2 = 25.

Hence, 4a2 + 9b2 = 25.

Question 4

If 3x - 4y = 16 and xy = 4, find the value of 9x2 + 16y2.

Answer

We know that,

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

9x2 + 16y2 = (3x)2 + (4y)2 = (3x - 4y)2 + 24xy.

Substituting values we get,

⇒ 9x2 + 16y2 = (16)2 + 24 × 4

⇒ 9x2 + 16y2 = 256 + 96

⇒ 9x2 + 16y2 = 352.

Hence, 9x2 + 16y2 = 352.

Question 5

If x + y = 8 and x - y = 2, find the value of 2x2 + 2y2.

Answer

We know that,

(x + y)2 = x2 + y2 + 2xy .....(i)

(x - y)2 = x2 + y2 - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)2 + (x - y)2 = x2 + x2 + y2 + y2 + 2xy - 2xy

= 2x2 + 2y2.

∴ 2x2 + 2y2 = (x + y)2 + (x - y)2.

Substituting values we get,

⇒ 2x2 + 2y2 = (8)2 + (2)2

⇒ 2x2 + 2y2 = 64 + 4 = 68.

Hence, 2x2 + 2y2 = 68.

Question 6

If a2 + b2 = 13 and ab = 6, find

(i) a + b

(ii) a - b

Answer

(i) We know that,

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

∴ (a + b) = a2+b2+2ab\sqrt{a^2 + b^2 + 2ab}

Substituting values we get,

⇒(a+b)=13+2×6⇒(a+b)=13+12⇒(a+b)=25⇒(a+b)=±5.\Rightarrow (a + b) = \sqrt{13 + 2 \times 6} \\[1em] \Rightarrow (a + b) = \sqrt{13 + 12} \\[1em] \Rightarrow (a + b) = \sqrt{25} \\[1em] \Rightarrow (a + b) = \pm 5.

Hence, a + b = ±5.

(ii) We know that,

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

∴ (a - b) = a2+b2−2ab\sqrt{a^2 + b^2 - 2ab}

Substituting values we get,

⇒(a−b)=13−2×6⇒(a−b)=13−12⇒(a−b)=1⇒(a−b)=±1.\Rightarrow (a - b) = \sqrt{13 - 2 \times 6} \\[1em] \Rightarrow (a - b) = \sqrt{13 - 12} \\[1em] \Rightarrow (a - b) = \sqrt{1} \\[1em] \Rightarrow (a - b) = \pm 1.

Hence, a - b = ±1.

Question 7

If a + b = 4 and ab = -12, find

(i) a - b

(ii) a2 - b2.

Answer

(i) We know that,

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

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

(a - b)2 = (a + b)2 - 4ab

(a - b) = (a+b)2−4ab\sqrt{(a + b)^2 - 4ab}

Substituting values we get,

⇒(a−b)=(4)2−4×(−12)⇒(a−b)=16+48⇒(a−b)=64⇒(a−b)=±8.\Rightarrow (a - b) = \sqrt{(4)^2 - 4 \times (-12)} \\[1em] \Rightarrow (a - b) = \sqrt{16 + 48} \\[1em] \Rightarrow (a - b) = \sqrt{64} \\[1em] \Rightarrow (a - b) = \pm 8.

Hence, a - b = ±8.

(ii) We know that,

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

Substituting values we get,

⇒ a2 - b2 = 4 × ±8 = ±32.

Hence, a2 - b2 = ±32.

Question 8

If p - q = 9 and pq = 36, evaluate

(i) p + q

(ii) p2 - q2.

Answer

(i) We know that,

(p - q)2 = p2 + q2 - 2pq

⇒ (p - q)2 = p2 + q2 + 2pq - 2pq - 2pq

⇒ (p - q)2 = (p + q)2 - 4pq

⇒ (p + q)2 = (p - q)2 + 4pq

⇒ (p + q) = (p−q)2+4pq\sqrt{(p - q)^2 + 4pq}

Substituting value we get,

⇒p+q=92+4×36⇒p+q=81+144⇒p+q=225⇒p+q=±15.\Rightarrow p + q = \sqrt{9^2 + 4 \times 36} \\[1em] \Rightarrow p + q = \sqrt{81 + 144} \\[1em] \Rightarrow p + q = \sqrt{225} \\[1em] \Rightarrow p + q = \pm 15.

Hence, p + q = ±15.

(ii) p2 - q2 = (p - q)(p + q).

Substituting value we get,

⇒ p2 - q2 = 9 × ±15 = ±135.

Hence, p2 - q2 = ±135.

Question 9

If x + y = 6 and x - y = 4, find

(i) x2 + y2

(ii) xy.

Answer

(i) We know that,

(x + y)2 = x2 + y2 + 2xy .....(i)

(x - y)2 = x2 + y2 - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)2 + (x - y)2 = x2 + x2 + y2 + y2 + 2xy - 2xy = 2x2 + 2y2.

⇒ 2x2 + 2y2 = (x + y)2 + (x - y)2.

∴ x2 + y2 = (x+y)2+(x−y)22\dfrac{(x + y)^2 + (x - y)^2}{2}

Substituting values we get,

⇒x2+y2=(6)2+(4)22=36+162=522=26.\Rightarrow x^2 + y^2 = \dfrac{(6)^2 + (4)^2}{2} \\[1em] = \dfrac{36 + 16}{2} \\[1em] = \dfrac{52}{2} \\[1em] = 26.

Hence, x2 + y2 = 26.

(ii) We know that,

(x + y)2 = x2 + y2 + 2xy .....(i)

(x - y)2 = x2 + y2 - 2xy ....(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)2 - (x - y)2 = x2 - x2 + y2 - y2 + 2xy - (-2xy) = 4xy.

⇒ (x + y)2 - (x - y)2 = 4xy.

∴ xy = (x+y)2−(x−y)24\dfrac{(x + y)^2 - (x - y)^2}{4}

Substituting values we get,

xy=62−424=36−164=204=5.xy = \dfrac{6^2 - 4^2}{4} \\[1em] = \dfrac{36 - 16}{4} \\[1em] = \dfrac{20}{4} \\[1em] = 5.

Hence, xy = 5.

Question 10

If x - 3 = 1x\dfrac{1}{x}, find the value of x2+1x2.x^2 + \dfrac{1}{x^2}.

Answer

Given,

∴x−3=1x∴x−1x=3\phantom{\therefore} x - 3 = \dfrac{1}{x} \\[1em] \therefore x - \dfrac{1}{x} = 3 \\[1em]

We know that,

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

Substituting values we get,

x2+1x2=(3)2−2=9−2=7.x^2 + \dfrac{1}{x^2} = (3)^2 - 2 \\[1em] = 9 - 2 \\[1em] = 7.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 7.

Question 11

If x + y = 8 and xy = 3343\dfrac{3}{4}, find the values of

(i) x - y

(ii) 3(x2 + y2)

(iii) 5(x2 + y2) + 4(x - y).

Answer

(i) We know that,

(x + y)2 = x2 + y2 + 2xy .....(i)

(x - y)2 = x2 + y2 - 2xy .....(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)2 - (x - y)2 = x2 - x2 + y2 - y2 + 2xy - (-2xy) = 4xy.

⇒ (x + y)2 - (x - y)2 = 4xy.

∴ (x - y) = (x+y)2−4xy\sqrt{(x + y)^2 - 4xy}.

Substituting values we get,

x−y=82−4×334=64−4×154=64−15=49=±7.x - y = \sqrt{8^2 - 4 \times 3\dfrac{3}{4}} \\[1em] = \sqrt{64 - 4 \times \dfrac{15}{4}} \\[1em] = \sqrt{64 - 15} \\[1em] = \sqrt{49} \\[1em] = \pm 7.

Hence, x - y = ±7.

(ii) We know that,

3(x2 + y2) = 3[(x + y)2 - 2xy].

Substituting values we get,

3(x2+y2)=3[(8)2−2×334]=3(64−2×154)=3(64−152)=3(128−152)=3×1132=3392=16912.3(x^2 + y^2) = 3\Big[(8)^2 - 2 \times 3\dfrac{3}{4}\Big] \\[1em] = 3\Big(64 - 2 \times \dfrac{15}{4}\Big) \\[1em] = 3\Big(64 - \dfrac{15}{2}\Big) \\[1em] = 3\Big(\dfrac{128 - 15}{2}\Big) \\[1em] = 3 \times \dfrac{113}{2} \\[1em] = \dfrac{339}{2} \\[1em] = 169\dfrac{1}{2}.

Hence, 3(x2 + y2) = 16912.169\dfrac{1}{2}.

(iii) From parts (i) and (ii) we get,

(x - y) = ±7 and x2 + y2 = 1132\dfrac{113}{2}.

When (x - y) = 7,

Substituting values we get,

5(x2+y2)+4(x−y)=5×1132+4×7=5652+28=565+562=6212=31012.5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times 7 \\[1em] = \dfrac{565}{2} + 28 \\[1em] = \dfrac{565 + 56}{2} \\[1em] = \dfrac{621}{2} \\[1em] = 310\dfrac{1}{2}.

When (x - y) = -7,

Substituting values we get,

5(x2+y2)+4(x−y)=5×1132+4×−7=5652−28=565−562=5092=25412.5(x^2 + y^2) + 4(x - y) = 5 \times \dfrac{113}{2} + 4 \times -7 \\[1em] = \dfrac{565}{2} - 28 \\[1em] = \dfrac{565 - 56}{2} \\[1em] = \dfrac{509}{2} \\[1em] = 254\dfrac{1}{2}.

Hence, 5(x2 + y2) + 4(x - y) = 31012 or 25412310\dfrac{1}{2} \text{ or } 254\dfrac{1}{2}.

Question 12

If x2 + y2 = 34 and xy = 101210\dfrac{1}{2}, find the value of 2(x + y)2 + (x - y)2.

Answer

We know that,

2(x+y)2+(x−y)2=2(x2+y2+2xy)+(x2+y2−2xy)=2x2+2y2+4xy+x2+y2−2xy=3x2+3y2+2xy=3(x2+y2)+2xy.2(x + y)^2 + (x - y)^2 = 2(x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) \\[1em] = 2x^2 + 2y^2 + 4xy + x^2 + y^2 - 2xy \\[1em] = 3x^2 + 3y^2 + 2xy \\[1em] = 3(x^2 + y^2) + 2xy.

Substituting values in above equation,

=3×34+2×1012=102+2×212=102+21=123.= 3 \times 34 + 2 \times 10\dfrac{1}{2} \\[1em] = 102 + 2 \times \dfrac{21}{2} \\[1em] = 102 + 21 \\[1em] = 123.

Hence, 2(x + y)2 + (x - y)2 = 123.

Question 13

If a - b = 3 and ab = 4, find a3 - b3.

Answer

We know that,

⇒ (a - b)3 = a3 - b3 - 3ab(a - b).

∴ a3 - b3 = (a - b)3 + 3ab(a - b).

Substituting values we get,

⇒ a3 - b3 = (3)3 + 3 × 4 × 3
⇒ a3 - b3 = 27 + 36 = 63.

Hence, a3 - b3 = 63.

Question 14

If 2a - 3b = 3 and ab = 2, find the value of 8a3 - 27b3.

Answer

We know that,

⇒ (a - b)3 = a3 - b3 - 3ab(a - b).

∴ a3 - b3 = (a - b)3 + 3ab(a - b).

∴ 8a3 - 27b3 = (2a)3 - (3b)3 = (2a - 3b)3 + 3 × 2a × 3b (2a - 3b)

Substituting values we get,

⇒ 8a3 - 27b3 = (2a - 3b)3 + 3 × 2a × 3b (2a - 3b)

⇒ 8a3 - 27b3 = 33 + 18ab × 3

⇒ 8a3 - 27b3 = 27 + 18 × 2 × 3

⇒ 8a3 - 27b3 = 27 + 108

⇒ 8a3 - 27b3 = 135.

Hence, 8a3 - 27b3 = 135.

Question 15

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

(i) x2+1x2x^2 + \dfrac{1}{x^2}

(ii) x4+1x4x^4 + \dfrac{1}{x^4}

(iii) x3+1x3x^3 + \dfrac{1}{x^3}

(iv) x−1xx - \dfrac{1}{x}.

Answer

(i) We know that,

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

Substituting values we get,

x2+1x2=42−2=16−2=14.x^2 + \dfrac{1}{x^2} = 4^2 - 2 \\[1em] = 16 - 2 \\[1em] = 14.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 14.

(ii) We know that,

⇒(x+1x)2=x2+1x2+2⇒x2+1x2=(x+1x)2−2∴x4+1x4=(x2)2+1(x2)2=(x2+1x2)2−2.\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2 \\[1em] \therefore x^4 + \dfrac{1}{x^4} = (x^2)^2 + \dfrac{1}{(x^2)^2} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2.

Substituting values we get,

x4+1x4=142−2=196−2=194.x^4 + \dfrac{1}{x^4} = 14^2 - 2 \\[1em] = 196 - 2 \\[1em] = 194.

Hence, x4+1x4x^4 + \dfrac{1}{x^4} = 194.

(iii) We know that,

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

Substituting values we get,

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

Hence, x3+1x3x^3 + \dfrac{1}{x^3} = 52.

(iv) We know that,

⇒(x−1x)2=x2+1x2−2∴x−1x=x2+1x2−2\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \therefore x - \dfrac{1}{x} = \sqrt{x^2 + \dfrac{1}{x^2} - 2}

Substituting values we get,

x−1x=14−2=12=4×3=±23.x - \dfrac{1}{x} = \sqrt{14 - 2} \\[1em] = \sqrt{12} \\[1em] = \sqrt{4 \times 3} \\[1em] = \pm 2\sqrt{3}.

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

Question 16

If x−1x=5x - \dfrac{1}{x} = 5, find the value of x4+1x4x^4 + \dfrac{1}{x^4}.

Answer

x4+1x4=(x2)2+(1x2)2.....[i](x2+1x2)2=(x2)2+2×(x2)2×1x2+(1x2)2∴(x2)2+(1x2)2=(x2+1x2)2−2x^4 + \dfrac{1}{x^4} = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 .....[i] \\[1em] \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + 2 \times (x^2)^2 \times \dfrac{1}{x^2} + \Big(\dfrac{1}{x^2}\Big)^2 \\[1em] \therefore (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 \\[1em]

Putting this value of (x2)2+(1x2)2(x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 in eqn (i), we get:

x4+1x4=(x2+1x2)2−2 .....[ii](x−1x)2=x2−2×x×1x+(1x)2∴x2+1x2=(x−1x)2+2x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2 \space .....[ii] \\[1em] \Big(x - \dfrac{1}{x}\Big)^2 = x^2 - 2 \times x \times \dfrac{1}{x} + \Big(\dfrac{1}{x}\Big)^2 \\[1em] \therefore x^2 + \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)^2 + 2 \\[1em]

Putting this value of x2+1x2x^2 + \dfrac{1}{x^2} in eqn (ii), we get:

x4+1x4=[(x−1x)2+2]2−2x^4 + \dfrac{1}{x^4} = \Big[\Big(x - \dfrac{1}{x}\Big)^2 + 2\Big]^2 - 2 \\[1em]

Putting the given values in above eqn, we get:

x4+1x4=(52+2)2−2=(27)2−2=729−2=727.x^4 + \dfrac{1}{x^4} = (5^2 + 2)^2 - 2 \\[1em] = (27)^2 - 2 \\[1em] = 729 - 2 \\[1em] = 727.

Hence, x4+1x4x^4 + \dfrac{1}{x^4} = 727.

Question 17

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

(i) x2+1x2x^2 + \dfrac{1}{x^2}

(ii) x+1xx + \dfrac{1}{x}

(iii) x3+1x3x^3 + \dfrac{1}{x^3}

Answer

(i) We know that,

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

Substituting values we get,

x2+1x2=(5)2+2=5+2=7.x^2 + \dfrac{1}{x^2} = (\sqrt{5})^2 + 2 \\[1em] = 5 + 2 \\[1em] = 7.

Hence, x2+1x2x^2 + \dfrac{1}{x^2} = 7.

(ii) We know that,

⇒(x+1x)2=x2+1x2+2∴x+1x=x2+1x2+2\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \therefore x + \dfrac{1}{x} = \sqrt{x^2 + \dfrac{1}{x^2} + 2}

Substituting values we get,

x+1x=7+2=9=±3.x + \dfrac{1}{x} = \sqrt{7 + 2} \\[1em] = \sqrt{9} \\[1em] = \pm 3.

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

(iii) We know that,

⇒(x+1x)3=x3+1x3+3(x+1x)∴x3+1x3=(x+1x)3−3(x+1x).\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \therefore x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big).

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

Substituting values we get,

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

When x+1x=−3x + \dfrac{1}{x} = -3.

Substituting values we get,

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

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

Question 18

If x+1xx + \dfrac{1}{x} = 6, find

(i) x−1xx - \dfrac{1}{x}

(ii) x2−1x2x^2 - \dfrac{1}{x^2}

Answer

(i) We know that,

⇒(x−1x)2=x2+1x2−2 ......(i)⇒(x+1x)2=x2+1x2+2 ......(ii)\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \text{ ......(i)} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \text{ ......(ii)}

Eq. (i) can be written as,

⇒(x−1x)2=x2+1x2+2−4⇒(x−1x)2=(x+1x)2−4∴x−1x=(x+1x)2−4\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = \Big(x + \dfrac{1}{x}\Big)^2 - 4 \\[1em] \therefore x - \dfrac{1}{x} = \sqrt{\Big(x + \dfrac{1}{x}\Big)^2 - 4}

Substituting values we get,

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

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

(ii) We know that,

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

When, x−1x=42x - \dfrac{1}{x} = 4\sqrt{2}

Substituting values we get,

x2−1x2=6×42=242.x^2 - \dfrac{1}{x^2} = 6 \times 4\sqrt{2} = 24\sqrt{2}.

When, x−1x=−42x - \dfrac{1}{x} = -4\sqrt{2}

Substituting values we get,

x2−1x2=6×−42=−242.x^2 - \dfrac{1}{x^2} = 6 \times -4\sqrt{2} = -24\sqrt{2}.

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

Question 19

If x+1x=2x + \dfrac{1}{x} = 2, prove that x2+1x2=x3+1x3=x4+1x4x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}.

Answer

We know that,

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

Substituting values we get,

x2+1x2=22−2=4−2=2x^2 + \dfrac{1}{x^2} = 2^2 - 2 = 4 - 2 = 2 ........(i)

We know that,

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

Substituting values we get,

x3+1x3=(2)3−3×2=8−6=2x^3 + \dfrac{1}{x^3} = (2)^3 - 3 \times 2 = 8 - 6 = 2 .......(ii)

We know that,

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

Substituting values we get,

x4+1x4=22−2=4−2=2x^4 + \dfrac{1}{x^4} = 2^2 - 2 = 4 - 2 = 2 ...........(iii)

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

Hence proved, x2+1x2=x3+1x3=x4+1x4.x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}. when x+1x=2x + \dfrac{1}{x} = 2

Question 20

If x−2x=3x - \dfrac{2}{x} = 3, find the value of x3−8x3x^3 - \dfrac{8}{x^3}.

Answer

We know that,

⇒ (a - b)3 = a3 - b3 - 3ab(a - b)

⇒ a3 - b3 = (a - b)3 + 3ab(a - b).

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

Substituting values we get,

x3−8x3=33+6×3=27+18=45.x^3 - \dfrac{8}{x^3} = 3^3 + 6 \times 3 = 27 + 18 = 45.

Hence, the value of x3−8x3=45.x^3 - \dfrac{8}{x^3} = 45.

Question 21

If a + 2b = 5, prove that a3 + 8b3 + 30ab = 125.

Answer

We know that,

⇒ (a + 2b)3 = a3 + 8b3 + 3(a)(2b)(a + 2b)

Substituting values in above formula,

⇒ 53 = a3 + 8b3 + 6ab × 5
⇒ 125 = a3 + 8b3 + 30ab.

Hence, proved that a3 + 8b3 + 30ab = 125.

Question 22

If a+1a=pa + \dfrac{1}{a} = p, prove that a3+1a3=p(p2−3)a^3 + \dfrac{1}{a^3} = p(p^2 - 3).

Answer

We know that,

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

Substituting values we get,

a3+1a3=p3−3p=p(p2−3).a^3 + \dfrac{1}{a^3} = p^3 - 3p \\[1em] = p(p^2 - 3).

Hence, proved that a3+1a3=p(p2−3)a^3 + \dfrac{1}{a^3} = p(p^2 - 3).

Question 24

If x2+1x2=27x^2 + \dfrac{1}{x^2} = 27, find the value of 3x3+5x−3x3−5x.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x}.

Answer

We know that,

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

Substituting values we get,

x−1x=27−2=25=±5.x - \dfrac{1}{x} = \sqrt{27 - 2} = \sqrt{25} = \pm 5.

Here, 3x3+5x−3x3−5x3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} can be written as,

3x3+5x−3x3−5x=3(x3−1x3)+5(x−1x)=3[(x−1x)3+3(x−1x)]+5(x−1x).3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3\Big(x^3 - \dfrac{1}{x^3}\Big) + 5\Big(x - \dfrac{1}{x}\Big) \\[1em] = 3\Big[\Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big)\Big] + 5\Big(x - \dfrac{1}{x}\Big).

Considering x−1x=5x - \dfrac{1}{x} = 5 in first case and substituting value we get,

3x3+5x−3x3−5x=3(53+3×5)+(5×5)=3(125+15)+25=(3×140)+25=420+25=445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3(5^3 + 3 \times 5) + (5 \times 5) \\[1em] = 3(125 + 15) + 25 \\[1em] = (3 \times 140) + 25 \\[1em] = 420 + 25 \\[1em] = 445.

Considering x−1x=−5x - \dfrac{1}{x} = -5 in second case and substituting value we get,

3x3+5x−3x3−5x=3[(−5)3+3×(−5)]+[5×(−5)]=3(−125−15)−25=(3×−140)−25=−420−25=−445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = 3[(-5)^3 + 3 \times (-5)] + [5 \times (-5)] \\[1em] = 3(-125 - 15) - 25 \\[1em] = (3 \times -140) - 25 \\[1em] = -420 - 25 \\[1em] = -445.

Hence, 3x3+5x−3x3−5x=±445.3x^3 + 5x - \dfrac{3}{x^3} - \dfrac{5}{x} = \pm 445.

Question 25

If x2+125x2=835, find x+15xx^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}, \text{ find } x + \dfrac{1}{5x}.

Answer

x2+125x2=(x)2+1(5x)2....[i](x+15x)2=(x)2+2×x×15x+1(5x)2=(x)2+1(5x)2+25∴x2+125x2=(x+15x)2−25x^2 + \dfrac{1}{25x^2} = (x)^2 + \dfrac{1}{(5x)^2} ....[i] \\[1em] \Big(x + \dfrac{1}{5x}\Big)^2 = (x)^2 + 2 \times x \times \dfrac{1}{5x} + \dfrac{1}{(5x)^2} \\[1em] = (x)^2 + \dfrac{1}{(5x)^2} + \dfrac{2}{5} \\[1em] \therefore x^2 + \dfrac{1}{25x^2} = \Big(x + \dfrac{1}{5x}\Big)^2 - \dfrac{2}{5}

Putting this value of x2+125x2x^2 + \dfrac{1}{25x^2} in eqn (i), we get:

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

Let x+15xx + \dfrac{1}{5x} be a.

Substituting values we get,

⇒835=a2−25⇒435=5a2−25⇒5a2−2=43⇒5a2=45⇒a2=9⇒a=±3.\Rightarrow 8\dfrac{3}{5} = a^2 - \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{43}{5} = \dfrac{5a^2 - 2}{5} \\[1em] \Rightarrow 5a^2 - 2 = 43 \\[1em] \Rightarrow 5a^2 = 45 \\[1em] \Rightarrow a^2 = 9 \\[1em] \Rightarrow a = \pm 3.

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

Question 26

If x2+14x2=8, find x3+18x3x^2 + \dfrac{1}{4x^2} = 8, \text{ find } x^3 + \dfrac{1}{8x^3}.

Answer

We know that,

⇒(x+12x)2=x2+14x2+2×x×12x⇒(x+12x)2=x2+14x2+1∴x2+14x2=(x+12x)2−1\Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = x^2 + \dfrac{1}{4x^2} + 2 \times x \times \dfrac{1}{2x} \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = x^2 + \dfrac{1}{4x^2} + 1 \\[1em] \therefore x^2 + \dfrac{1}{4x^2} = \Big(x + \dfrac{1}{2x}\Big)^2 - 1

Substituting values in above equation we get,

⇒8=(x+12x)2−1⇒(x+12x)2=8+1⇒(x+12x)2=9⇒x+12x=9⇒x+12x=±3.\Rightarrow 8 = \Big(x + \dfrac{1}{2x}\Big)^2 - 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 8 + 1 \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^2 = 9 \\[1em] \Rightarrow x + \dfrac{1}{2x} = \sqrt{9} \\[1em] \Rightarrow x + \dfrac{1}{2x} = \pm 3.

We know that,

⇒(a+b)3=a3+b3+3ab(a+b)⇒(x+12x)3=x3+(12x)3+3×x×12x(x+12x)⇒(x+12x)3=x3+18x3+32(x+12x)∴x3+18x3=(x+12x)3−32(x+12x).\Rightarrow (a + b)^3 = a^3 + b^3 + 3ab(a + b) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \Big(\dfrac{1}{2x}\Big)^3 + 3 \times x \times \dfrac{1}{2x}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{2x}\Big)^3 = x^3 + \dfrac{1}{8x^3} + \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big) \\[1em] \therefore x^3 + \dfrac{1}{8x^3} = \Big(x + \dfrac{1}{2x}\Big)^3 - \dfrac{3}{2}\Big(x + \dfrac{1}{2x}\Big)..

Case 1: x+12x=3x + \dfrac{1}{2x} = 3 substituting values we get,

x3+18x3=33−32×3=27−92=54−92=452.x^3 + \dfrac{1}{8x^3} = 3^3 - \dfrac{3}{2} \times 3 \\[1em] = 27 - \dfrac{9}{2} \\[1em] = \dfrac{54 - 9}{2} \\[1em] = \dfrac{45}{2}.

Case 2 : x+12x=−3x + \dfrac{1}{2x} = -3, substituting values we get,

x3+18x3=(−3)3−32×(−3)=−27+92=−54+92=−452.x^3 + \dfrac{1}{8x^3} = (-3)^3 - \dfrac{3}{2} \times (-3) \\[1em] = -27 + \dfrac{9}{2} \\[1em] = \dfrac{-54 + 9}{2} \\[1em] = -\dfrac{45}{2}.

Hence, x3+18x3=±452=±2212x^3 + \dfrac{1}{8x^3} = \pm \dfrac{45}{2} = \pm 22\dfrac{1}{2}.

Question 27

If a2 - 3a + 1 = 0, find

(i) a2+1a2a^2 + \dfrac{1}{a^2}

(ii) a3+1a3a^3 + \dfrac{1}{a^3}

Answer

Dividing each term of a2 - 3a + 1 = 0 by a we get,

a+1a=3a + \dfrac{1}{a} = 3.

(i) We know that,

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

Substituting values we get,

a2+1a2=32−2=9−2=7.a^2 + \dfrac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7.

Hence, a2+1a2a^2 + \dfrac{1}{a^2} = 7.

(ii) We know that,

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

Substituting values we get,

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

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

Question 28

If a = 1a−5\dfrac{1}{a - 5}, find

(i) a−1aa - \dfrac{1}{a}

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

(iii) a2−1a2a^2 - \dfrac{1}{a^2}

Answer

Given,

⇒a=1a−5∴a(a−5)=1⇒a2−5a=1⇒a2−5a−1=0\Rightarrow a = \dfrac{1}{a - 5} \\[1em] \therefore a(a - 5) = 1 \\[1em] \Rightarrow a^2 - 5a = 1 \\[1em] \Rightarrow a^2 - 5a - 1 = 0

Now,

(i) Dividing above equation by a we get,

⇒a−5−1a=0⇒a−1a=5.\Rightarrow a - 5 - \dfrac{1}{a} = 0 \\[1em] \Rightarrow a - \dfrac{1}{a} = 5..

Hence, a−1a=5a - \dfrac{1}{a} = 5.

(ii) We know that,

⇒(a+1a)2=a2+1a2+2 .....(i)⇒(a−1a)2=a2+1a2−2 .....(ii)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \space .....(i) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \space .....(ii) \\[1em]

Subtracting eq. (ii) from (i) we get,

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

Substituting values we get,

a+1a=52+4=25+4=±29.a + \dfrac{1}{a} = \sqrt{5^2 + 4} \\[1em] = \sqrt{25 + 4} \\[1em] = \pm \sqrt{29}.

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

(iii) We know that,

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

Substituting values we get,

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

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

Question 29

If (x+1x)2=3, find x3+1x3.\Big(x + \dfrac{1}{x}\Big)^2 = 3, \text{ find } x^3 + \dfrac{1}{x^3}.

Answer

Given,

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

We know that,

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

When x+1x=3x + \dfrac{1}{x} = \sqrt 3 substituting values we get,

x3+1x3=(3)3−3×3=33−33=0.x^3 + \dfrac{1}{x^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0.

When x+1x=−3x + \dfrac{1}{x} = -\sqrt{3} substituting values we get,

x3+1x3=(−3)3−3×(−3)=−33+33=0.x^3 + \dfrac{1}{x^3} = (-\sqrt{3})^3 - 3 \times (-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0.

Hence, x3+1x3=0x^3 + \dfrac{1}{x^3} = 0.

Question 30

If x=5−26x = 5 - 2\sqrt{6}, find the value of x+1x\sqrt{x} + \dfrac{1}{\sqrt{x}}

Answer

Given,

x=5−26⇒1x=15−26⇒1x=15−26×5+265+26⇒1x=5+26(5−26)(5+26)⇒1x=5+2652−(26)2⇒1x=5+2625−24⇒1x=5+261=5+26.x = 5 - 2\sqrt{6} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{5- 2\sqrt{6}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} \times \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{25 - 24} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6}.

So,

x+1x=5−26+5+26=10.x + \dfrac{1}{x} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6} = 10.

We know that,

⇒(x+1x)2=x+1x+2×x×1x⇒(x+1x)2=x+1x+2⇒x+1x=x+1x+2=10+2=12=±23.\Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \times \sqrt{x} \times \dfrac{1}{\sqrt{x}} \\[1em] \Rightarrow \Big(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Big)^2 = x + \dfrac{1}{x} + 2 \\[1em] \Rightarrow \sqrt{x} + \dfrac{1}{\sqrt{x}} = \sqrt{x + \dfrac{1}{x} + 2} = \sqrt{10 + 2} = \sqrt{12} = \pm 2\sqrt{3}.

Hence, x+1x=±23.\sqrt{x} + \dfrac{1}{\sqrt{x}} = \pm 2\sqrt{3}.

Question 31

If a + b + c = 12 and ab + bc + ca = 22, find a2 + b2 + c2.

Answer

We know that,

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

Substituting values we get,

a2 + b2 + c2 = (12)2 - 2(22) = 144 - 44 = 100.

Hence, a2 + b2 + c2 = 100.

Question 32

If a + b + c = 12 and a2 + b2 + c2 = 100, find ab + bc + ca.

Answer

We know that,

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

Substituting values we get,

⇒ 100 = 122 - 2(ab + bc + ca)

⇒ 100 = 144 - 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 144 - 100

⇒ 2(ab + bc + ca) = 44

⇒ (ab + bc + ca) = 22.

Hence, ab + bc + ca = 22.

Question 33

If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.

Answer

We know that,

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

Substituting values we get,

⇒ (a + b + c)2 = 125 + 2(50) = 125 + 100 = 225.

∴(a+b+c)=225=±15.\therefore (a + b + c) = \sqrt{225} = \pm 15.

Hence, (a + b + c) = ± 15.

Question 34

If a + b - c = 5 and a2 + b2 + c2 = 29, find the value of ab - bc - ca.

Answer

Given, a + b - c = 5. Squaring both sides we get,

⇒ (a + b - c)2 = 52

⇒ a2 + b2 + c2 + 2ab - 2bc - 2ca = 25

Substituting values we get,

⇒ 29 + 2(ab - bc - ca) = 25

⇒ 2(ab - bc - ca) = 25 - 29

⇒ 2(ab - bc - ca) = -4

⇒ (ab - bc - ca) = -2.

Hence, (ab - bc - ca) = -2.

Question 35

If a - b = 7 and a2 + b2 = 85, then find the value of a3 - b3.

Answer

We know that,

⇒ (a - b)2 = a2 + b2 - 2ab

⇒ (7)2 = 85 - 2ab

⇒ 49 = 85 - 2ab

⇒ 2ab = 85 - 49

⇒ 2ab = 36

⇒ ab = 18.

We know that,

a3 - b3 = (a - b)(a2 + b2 + ab)

Substituting values we get,

⇒ a3 - b3 = 7(85 + 18) = 7(103) = 721.

Hence, a3 - b3 = 721.

Question 36

If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.

Answer

Given,

x = y - 3 or x - y = -3 and

x2 + y2 = 29.

We know that,

⇒ (x + y)2 = x2 + y2 + 2xy ......(i)

⇒ (x - y)2 = x2 + y2 - 2xy .......(ii)

Subtracting eq. (ii) from (i) we get,

⇒ (x + y)2 - (x - y)2 = x2 + y2 + 2xy - (x2 + y2 - 2xy)

⇒ (x + y)2 - (x - y)2 = 4xy

⇒ x2 + y2 = (x - y)2 + 4xy

Substituting values we get,

⇒ 29 = (-3)2 + 4xy
⇒ 29 = 9 + 4xy
⇒ 29 - 9 = 4xy
⇒ 4xy = 20
⇒ xy = 5.

Hence, xy = 5.

Question 37

If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.

Answer

Let two numbers be x and y.

Given,

x + y = 8 and xy = 15.

We know that,

⇒ x3 + y3 = (x + y)3 - 3xy(x + y)

⇒ x3 + y3 = 83 - 3(15)(8)

⇒ x3 + y3 = 512 - 360

⇒ x3 + y3 = 152.

Hence, x3 + y3 = 152.

Multiple Choice Questions

Question 1

If x+1x=2, then x2+1x2x + \dfrac{1}{x} = 2, \text{ then } x^2 + \dfrac{1}{x^2} is equal to

  1. 4

  2. 2

  3. 0

  4. none of these

Answer

We know that,

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

Substituting values we get,

x2+1x2=(2)2−2=4−2=2x^2 + \dfrac{1}{x^2} = (2)^2 - 2 = 4 - 2 = 2.

Hence, Option 2 is the correct option.

Question 2

If x2 + y2 = 9 and xy = 8, then x + y is equal to

  1. 25

  2. 5

  3. -5

  4. ±5

Answer

We know that,

(x+y)=x2+y2+2xy(x + y) = \sqrt{x^2 + y^2 + 2xy}.

Substituting values we get,

(x+y)=9+2×8=9+16=25=±5(x + y) = \sqrt{9 + 2 \times 8} = \sqrt{9 + 16} = \sqrt{25} = \pm 5.

Hence, Option 4 is the correct option.

Question 3

(102)2 - (98)2 is equal to

  1. 200

  2. 400

  3. 600

  4. 800

Answer

Applying a2 - b2 = (a - b)(a + b) formula to (102)2 - (98)2 we get,

(102)2 - (98)2 = (102 - 98)(102 + 98) = 4 × 200 = 800.

Hence, Option 4 is the correct option.

Question 4

96 × 104 is equal to

  1. 9984

  2. 9974

  3. 9964

  4. none of these

Answer

We can write 96 × 104 as (100 - 4) × (100 + 4).

By a2 - b2 = (a - b)(a + b) formula,

(100 - 4) × (100 + 4) = (100)2 - (4)2 = 10000 - 16 = 9984.

Hence, Option 1 is the correct option.

Question 5

1032−972200\dfrac{103^2 - 97^2}{200} is equal to

  1. 3

  2. 4

  3. 5

  4. 6

Answer

By a2 - b2 = (a - b)(a + b) formula,

1032−972200=(103−97)(103+97)200=6×200200=6.\dfrac{103^2 - 97^2}{200} = \dfrac{(103 - 97)(103 + 97)}{200} \\[1em] = \dfrac{6 \times 200}{200} \\[1em] = 6.

Hence, Option 4 is the correct option.

Question 6

If x + y = 11 and xy = 24, then x2 + y2 is equal to

  1. 121

  2. 73

  3. 48

  4. 169

Answer

We know that,

x2 + y2 = (x + y)2 - 2xy.

Substituting values we get,

x2 + y2 = (11)2 - 2 × 24 = 121 - 48 = 73.

Hence, Option 2 is the correct option.

Question 7

The value of 2492 - 2482 is

  1. 12

  2. 477

  3. 487

  4. 497

Answer

By a2 - b2 = (a - b)(a + b) formula,

2492 - 2482 = (249 - 248)(249 + 248) = 1 × 497 = 497.

Hence, Option 4 is the correct option.

Question 8

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

  1. 1

  2. -1

  3. 0

  4. 12\dfrac{1}{2}

Answer

Given,

⇒xy+yx=−1⇒x2+y2xy=−1⇒x2+y2=−xy⇒x2+y2+xy=0.\Rightarrow \dfrac{x}{y} + \dfrac{y}{x} = -1 \\[1em] \Rightarrow \dfrac{x^2 + y^2}{xy} = -1 \\[1em] \Rightarrow x^2 + y^2 = -xy \\[1em] \Rightarrow x^2 + y^2 + xy = 0.

We know that,

x3 - y3 = (x - y)(x2 + y2 + xy) = (x - y) × 0 = 0.

Hence, Option 3 is the correct option.

Question 9

If a + b + c = 0, then the value of a3 + b3 + c3 is

  1. 0

  2. abc

  3. 2abc

  4. 3abc

Answer

We know that,

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

Substituting values we get,

a3 + b3 + c3 - 3abc = 0 × (a2 + b2 + c2 - bc - ca - ab)

a3 + b3 + c3 - 3abc = 0

a3 + b3 + c3 = 3abc.

Hence, Option 4 is the correct option.

Question 10

If x−2x=3, then x3−8x3x - \dfrac{2}{x} = 3, \text{ then } x^3 - \dfrac{8}{x^3} is equal to

  1. 27

  2. 36

  3. 45

  4. 54

Answer

We know that,

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

⇒ a3 - b3 = (a - b)3 + 3ab(a - b)

∴x3−8x3=(x−2x)3+6(x−2x)\therefore x^3 - \dfrac{8}{x^3} = \Big(x - \dfrac{2}{x}\Big)^3 + 6\Big(x - \dfrac{2}{x}\Big)

Substituting values we get,

x3−8x3=33+6×3=27+18=45.x^3 - \dfrac{8}{x^3} = 3^3 + 6 \times 3 \\[1em] = 27 + 18 \\[1em] = 45.

Hence, Option 3 is the correct option.

Question 11

Consider the following two statements :

Statement 1: If a + b = 0, then a2 + b2 = 0

Statement 2: a2 + b2 = (a + b)2.

Which of the following is valid?

  1. Both the statements are true.

  2. Both the statements are false.

  3. Statement 1 is true, and Statement 2 is false.

  4. Statement 1 is false, and Statement 2 is true.

Answer

Given,

⇒ a + b = 0

Squaring both sides, we get

⇒ (a + b)2 = 02

⇒ a2 + b2 + 2ab = 0

⇒ a2 + b2 = -2ab

∴ Statement 1 is false.

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

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.

Assertion Reason Type Questions

Question 1

Assertion (A): If x = 1x\dfrac{1}{x}, then x = ± 1.

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

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given,

⇒ x = 1x\dfrac{1}{x}

⇒ x.x = 1

⇒ x2 = 1

⇒ x = 1\sqrt{1}

⇒ x = ± 1

∴ Assertion (A) is true.

Solving,

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

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Question 2

Assertion (A): 1003 x 997 = 999991

Reason (R): (a - b)(a + b) = a2 - b2

  1. Assertion (A) is true, Reason (R) is false.

  2. Assertion (A) is false, Reason (R) is true.

  3. Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

  4. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Answer

Given,

⇒ (a - b)(a + b)

⇒ a(a + b) - b(a + b)

⇒ a2 + ab - ab - b2

⇒ a2 - b2

∴ Reason (R) is true.

Solving,

⇒ 1003 x 997

⇒ (1000 + 3)(1000 - 3)

⇒ 10002 - 32

⇒ 1000000 - 9

⇒ 999991.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Chapter Test

Question 1

Find the expansions of the following :

(i) (2x + 3y + 5)(2x + 3y - 5)

(ii) (6 - 4a - 7b)2

(iii) (7 - 3xy)3

(iv) (x + y + 2)3

Answer

(i) On solving,

⇒ (2x + 3y + 5)(2x + 3y - 5) = (2x + 3y)2 - 52

⇒ (2x + 3y + 5)(2x + 3y - 5) = 4x2 + 9y2 + 12xy - 25.

Hence, (2x + 3y + 5)(2x + 3y - 5) = 4x2 + 9y2 + 12xy - 25.

(ii) On solving,

⇒ (6 - 4a - 7b)2 = (6 - 4a)2 + (7b)2 - 2(6 - 4a)(7b)

⇒ (6 - 4a - 7b)2 = 36 + 16a2 - 48a + 49b2 - 14b(6 - 4a)

⇒ (6 - 4a - 7b)2 = 36 + 16a2 + 49b2 - 48a + 56ab - 84b.

Hence, (6 - 4a - 7b)2 = 36 + 16a2 + 49b2 - 48a + 56ab - 84b.

(iii) On solving,

⇒ (7 - 3xy)3 = (7)3 - (3xy)3 - 3(7)(3xy)(7 - 3xy)

⇒ (7 - 3xy)3 = 343 - 27x3y3 - 63xy(7 - 3xy)

⇒ (7 - 3xy)3 = 343 - 27x3y3 - 441xy + 189x2y2.

Hence, (7 - 3xy)3 = 343 - 27x3y3 - 441xy + 189x2y2.

(iv) On solving,

⇒ (x + y + 2)3 = (x)3 + (y + 2)3 + 3(x)(y + 2)(x + y + 2)

⇒ (x + y + 2)3 = x3 + y3 + 23 + 3(y)(2)(y + 2) + (3xy + 6x)(x + y + 2)

⇒ (x + y + 2)3 = x3 + y3 + 8 + 6y(y + 2) + 3xy(x + y + 2) + 6x(x + y + 2)

⇒ (x + y + 2)3 = x3 + y3 + 8 + 6y2 + 12y + 3x2y + 3xy2 + 6xy + 6x2 + 6xy + 12x

⇒ (x + y + 2)3 = x3 + y3 + 3x2y + 3xy2 + 6x2 + 6y2 + 12xy + 12x + 12y + 8.

Hence, (x + y + 2)3 = x3 + y3 + 3x2y + 3xy2 + 6x2 + 6y2 + 12xy + 12x + 12y + 8.

Question 2

Simplify (x - 2)(x + 2)(x2 + 4)(x4 + 16).

Answer

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

On solving,

(x−2)(x+2)(x2+4)(x2+16)=(x2−4)(x2+4)(x4+16)=(x4−16)(x4+16)=((x4)2−162)=x8−256.(x - 2)(x + 2)(x^2 + 4)(x^2 + 16) = (x^2 - 4)(x^2 + 4)(x^4 + 16) \\[1em] = (x^4 - 16)(x^4 + 16) \\[1em] = ((x^4)^2 - 16^2) \\[1em] = x^8 - 256.

On simplifying we get, (x - 2)(x + 2)(x2 + 4)(x4 + 16) = x8 - 256.

Question 3

Evaluate 1002 × 998 by using a special product.

Answer

We can write 1002 × 998 as (1000 + 2)(1000 - 2).

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

We get,

(1000 + 2)(1000 - 2) = 10002 - 22 = 1000000 - 4 = 999996.

Hence, 1002 × 998 = 999996.

Question 4

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

Answer

Given,

a + 2b + 3c = 0
a + 2b = -3c.

Cubing both sides,

(a + 2b)3 = (-3c)3

a3 + (2b)3 + 3(a)(2b)(a + 2b) = -27c3

a3 + 8b3 + 6ab(-3c) = -27c3

a3 + 8b3 - 18abc = -27c3

a3 + 8b3 + 27c3 = 18abc.

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

Question 5

If 2x = 3y - 5, then find the value of 8x3 - 27y3 - 90xy + 125.

Answer

Given, 2x = 3y - 5 or 2x - 3y = -5

Cubing both sides we get,

⇒ (2x - 3y)3 = (-5)3

(2x)3 - (3y)3 - 3(2x)(3y)(2x - 3y) = -125

⇒ 8x3 - 27y3 - 18xy(2x - 3y) = -125

⇒ 8x3 - 27y3 - 18xy(-5) = -125

⇒ 8x3 - 27y3 + 90xy = -125

⇒ 8x3 - 27y3 + 90xy + 125 = 0.

Hence, 8x3 - 27y3 + 90xy + 125 = 0.

Question 6

If a2−1a2=5, evaluate a4+1a4a^2 - \dfrac{1}{a^2} = 5, \text{ evaluate } a^4 + \dfrac{1}{a^4}.

Answer

We know that,

a4+1a4=(a2−1a2)2+2a^4 + \dfrac{1}{a^4} = \Big(a^2 - \dfrac{1}{a^2}\Big)^2 + 2

Substituting values we get,

a4+1a4=(5)2+2=25+2=27.a^4 + \dfrac{1}{a^4} = (5)^2 + 2 = 25 + 2 = 27.

Hence, a4+1a4=27.a^4 + \dfrac{1}{a^4} = 27.

Question 7

If a+1a=p and a−1a=qa + \dfrac{1}{a} = p \text{ and } a - \dfrac{1}{a} = q, find the relation between p and q.

Answer

We know that,

⇒(a+1a)2=a2+1a2+2 .....(i)⇒(a−1a)2=a2+1a2−2 .....(ii)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \space .....(i)\\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \space .....(ii)

Subtracting eq. (ii) from (i) we get,

⇒(a+1a)2−(a−1a)2=4Substituting values we get,⇒p2−q2=4.\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \text{Substituting values we get}, \\[1em] \Rightarrow p^2 - q^2 = 4.

Hence, p2 - q2 = 4.

Question 8

If a2+1a=4\dfrac{a^2 + 1}{a} = 4, find the value of 2a3+2a32a^3 + \dfrac{2}{a^3}.

Answer

Given,

∴a2+1a=4∴a+1a=4.\phantom{\therefore} \dfrac{a^2 + 1}{a} = 4 \\[1em] \therefore a + \dfrac{1}{a} = 4.

Solving,

⇒2a3+2a3=2(a3+1a3)=2[(a+1a)3−3(a+1a)]=2[43−3×4]=2[64−12]=2×52=104.\Rightarrow 2a^3 + \dfrac{2}{a^3} = 2\Big(a^3 + \dfrac{1}{a^3}\Big) \\[1em] = 2\Big[\Big(a + \dfrac{1}{a}\Big)^3 - 3\Big(a + \dfrac{1}{a}\Big)\Big] \\[1em] = 2[4^3 - 3 \times 4] \\[1em] = 2[64 - 12] \\[1em] = 2 \times 52 \\[1em] = 104.

Hence, the value of 2a3+2a32a^3 + \dfrac{2}{a^3} = 104.

Question 9

If x=14−xx = \dfrac{1}{4 - x}, find the values of

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

(ii) x3+1x3x^3 + \dfrac{1}{x^3}

(iii) x6+1x6x^6 + \dfrac{1}{x^6}

Answer

(i) Given,

⇒x=14−x⇒x(4−x)=1⇒4x−x2=1\phantom{\Rightarrow} x = \dfrac{1}{4 - x} \\[1em] \Rightarrow x(4 - x) = 1 \\[1em] \Rightarrow 4x - x^2 = 1 \\[1em]

On dividing above equation by x,

⇒4−x=1x⇒x+1x=4.\Rightarrow 4 - x = \dfrac{1}{x} \\[1em] \Rightarrow x + \dfrac{1}{x} = 4.

Hence, the value of x+1xx + \dfrac{1}{x} = 4.

(ii) We know that,

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

Hence, the value of x3+1x3x^3 + \dfrac{1}{x^3} = 52.

(iii) We know,

(x6+1x6)=(x3+1x3)2−2\Big(x^6 + \dfrac{1}{x^6}\Big) = \Big(x^3 + \dfrac{1}{x^3}\Big)^2 - 2.

Substituting values we get,

(x6+1x6)=522−2=2704−2=2702.\Big(x^6 + \dfrac{1}{x^6}\Big) = 52^2 - 2 = 2704 - 2 = 2702.

Hence, the value of x6+1x6x^6 + \dfrac{1}{x^6} = 2702.

Question 10

If x−1x=3+22x - \dfrac{1}{x} = 3 + 2\sqrt{2}, find the value of 14(x3−1x3)\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big).

Answer

We know that,

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

Substituting values we get,

x3−1x3=(3+22)3+3(3+22)=(3)3+(22)3+3(3)(22)(3+22)+9+62=27+162+182(3+22)+9+62=27+9+162+542+72+62=108+762.x^3 - \dfrac{1}{x^3} = (3 + 2\sqrt{2})^3 + 3(3 + 2\sqrt{2}) \\[1em] = (3)^3 + (2\sqrt{2})^3 + 3(3)(2\sqrt{2})(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 16\sqrt{2} + 18\sqrt{2}(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 9 + 16\sqrt{2} + 54\sqrt{2} + 72 + 6\sqrt{2} \\[1em] = 108 + 76\sqrt{2}.

So, 14(x3−1x3)=14×(108+762)=27+192\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = \dfrac{1}{4} \times (108 + 76\sqrt{2}) = 27 + 19\sqrt{2}.

Hence, 14(x3−1x3)=27+192\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = 27 + 19\sqrt{2}.

Question 11

If x+1x=313x + \dfrac{1}{x} = 3\dfrac{1}{3}, find the value of x3−1x3x^3 - \dfrac{1}{x^3}.

Answer

We know that,

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

Substituting values we get,

x−1x=(313)2−4=(103)2−4=1009−4=100−369=649=±83.x - \dfrac{1}{x} = \sqrt{\Big(3\dfrac{1}{3}\Big)^2 - 4} \\[1em] = \sqrt{\Big(\dfrac{10}{3}\Big)^2 - 4} \\[1em] = \sqrt{\dfrac{100}{9} - 4} \\[1em] = \sqrt{\dfrac{100 - 36}{9}} \\[1em] = \sqrt{\dfrac{64}{9}} \\[1em] = \pm \dfrac{8}{3}.

We know that,

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

Substituting values we get,

x3−1x3=(±83)3+3×±(83)=±51227±8=±512±21627=±72827=±262627.x^3 - \dfrac{1}{x^3} = \Big(\pm \dfrac{8}{3}\Big)^3 + 3 \times \pm \Big(\dfrac{8}{3}\Big) \\[1em] = \pm \dfrac{512}{27} \pm 8 \\[1em] = \dfrac{\pm 512 \pm 216}{27} \\[1em] = \pm \dfrac{728}{27} \\[1em] = \pm 26\dfrac{26}{27}.

Hence, the value of x3−1x3=±262627x^3 - \dfrac{1}{x^3} = \pm 26\dfrac{26}{27}.

Question 12

If x=2−3x = 2 - \sqrt{3} then find the value of x3−1x3x^3 - \dfrac{1}{x^3}.

Answer

Given,

⇒x=2−3∴1x=12−3⇒1x=12−3×2+32+3⇒1x=2+34−(3)2⇒1x=2+34−3=2+3.\Rightarrow x = 2 - \sqrt{3} \\[1em] \therefore \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3}} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}.

So,

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

Cubing both sides we get,

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

Hence, x3−1x3=−303x^3 - \dfrac{1}{x^3} = -30\sqrt{3}.

Question 13

If the sum of two numbers is 7 and sum of their cubes is 133, find the sum of their squares.

Answer

Let the two numbers be x and y.

Given,

Sum of two numbers is 7.

∴ x + y = 7

Sum of cubes of two numbers is 133.

∴ x3 + y3 = 133

By formula,

⇒ (x + y)3 = x3 + y3 + 3xy(x + y)

Substituting values we get :

⇒ 73 = 133 + 3xy × 7

⇒ 343 = 133 + 21xy

⇒ 21xy = 343 - 133

⇒ 21xy = 210

⇒ xy = 21021\dfrac{210}{21} = 10.

By formula,

⇒ (x + y)2 = x2 + y2 + 2xy

Substituting values we get :

⇒ 72 = x2 + y2 + 2 × 10

⇒ 49 = x2 + y2 + 20

⇒ x2 + y2 = 49 - 20 = 29.

Hence, sum of the squares of the numbers = 29.

Question 14

If a - b = 7 and a3 - b3 = 133, find

(i) ab

(ii) a2 + b2.

Answer

(i) Given,

a - b = 7, cubing both sides we get,

⇒ (a - b)3 = 73

⇒ a3 - b3 - 3ab(a - b) = 343

⇒ 133 - 3ab(7) = 343

⇒ 133 - 21ab = 343

⇒ -21ab = 343 - 133

⇒ -21ab = 210

⇒ ab = -10.

Hence, ab = -10.

(ii) We know that,

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

Substituting values we get,

⇒ a2 + b2 = (7)2 + 2(-10)

⇒ a2 + b2 = 49 - 20 = 29.

Hence, a2 + b2 = 29.

Question 15

Find the coefficient of x2 in the expansion of

(x2 + x + 1)2 + (x2 - x + 1)2.

Answer

The above equation can be written as,

(x2+x+1)2+(x2−x+1)2=(x2+1)+x2+(x2+1)−x2=2(x2+1)2+x2=2(x2)2+1+2x2+x2=2x4+3x2+1=2x4+6x2+2.(x^2 + x + 1)^2 + (x^2 - x + 1)^2 = {(x^2 + 1) + x}^2 + {(x^2 + 1) - x}^2 \\[1em] = 2{(x^2 + 1)^2 + x^2} \\[1em] = 2{(x^2)^2 + 1 + 2x^2 + x^2} \\[1em] = 2{x^4 + 3x^2 + 1} \\[1em] = 2x^4 + 6x^2 + 2.

Hence, coefficient of x2 = 6.

Question 16

If x2 + y2 + z2 = xy + yz + zx, prove that x = y = z

Answer

Given,

x2 + y2 + z2 = xy + yz + zx

⇒ x2 + y2 + z2 - (xy + yz + zx) = 0

⇒ x2 + y2 + z2 - xy - yz - zx = 0

Multiplying by 2 on both sides,

⇒ 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx = 0

⇒ (x2 + y2 - 2xy) + (y2 + z2 - 2yz) + (z2 + x2 - 2xz) = 0

⇒ (x - y)2 + (y - z)2 + (z - x)2 = 0

⇒ x - y = 0, y - z = 0 and z - x = 0

⇒ x = y, y = z and z = x

⇒ x = y = z

Hence, proved that if x2 + y2 + z2 = xy + yz + zx, then x = y = z

PrevNext