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Chapter 12

Identities

Class - 8 Concise Mathematics Selina

Exercise 12(A)

Question 1(i)

(x + 2y)2 + (x - 2y)2 is equal to:

  1. 2x2 + 8y2 + 8xy

  2. x2 + 4y2

  3. 2x2 + 8y2

  4. 2x2 - 8y2

Answer

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

And using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

and,

[∵(x - y)2 = x2 - 2xy + y2]

= [x2 + 2 ×\times x ×\times 2y + (2y)2] + [x2 - 2 ×\times x ×\times 2y + (2y)2]

= [x2 + 4xy + 4y2] + [x2 - 4xy + 4y2]

= x2 + 4xy + 4y2 + x2 - 4xy + 4y2

= (x2 + x2) + (4xy - 4xy) + (4y2 + 4y2)

= 2x2 + 8y2

Hence, option 3 is the correct option.

Question 1(ii)

(a + b) (a - b) + (b - c) (b + c) + (c + a) (c - a) is equal to:

  1. 2a2 + 2b2 + 2c2

  2. a2 + b2 + c2 - 2ab - 2bc - 2ca

  3. 0

  4. none of these

Answer

[(a + b) (a - b)] + [(b - c) (b + c)] + [(c + a) (c - a)]

Using the formula,

[∵ (x + y)(x - y) = x2 - y2]

= [a2 - b2] + [b2 - c2] + [c2 - a2]

= a2 - b2 + b2 - c2 + c2 - a2

= (a2 - a2) + (- b2 + b2) + (- c2 + c2)

= 0

Hence, option 3 is the correct option.

Question 1(iii)

(3x - 4y)2 - (3x + 4y)2 is equal to:

  1. 18x2 + 32y2

  2. 18x2 - 32y2

  3. -48xy

  4. 48xy

Answer

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

Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

and,

[∵(x - y)2 = x2 - 2xy + y2]

= [(3x)2 - 2 ×\times (3x) ×\times (4y) + (4y)2] - [(3x)2 + 2 ×\times (3x) ×\times (4y) + (4y)2]

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

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

= (9x2 - 9x2) + (- 24xy - 24xy) + (16y2 - 16y2)

= - 48xy

Hence, option 3 is the correct option.

Question 1(iv)

The value of (0.8)2 - 0.32 + (0.2)2 is equal to:

  1. 1

  2. 3.6

  3. 0.36

  4. 0.036

Answer

(0.8)2 - 0.32 + (0.2)2

= (0.8)2 - 2 ×\times (0.8) ×\times (0.2) + (0.2)2

Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

= [0.8 - 0.2]2

= 0.62

= 0.36

Hence, option 3 is the correct option.

Question 1(v)

The value of (a - b - c)(a - b + c) is:

  1. a2 + b2 + c2 + 2ab

  2. a2 + b2 - c2 - 2ab

  3. a2 + b2 + c2 - 2ab

  4. a2 + b2 - c2 + 2ab

Answer

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

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

= a(1+1) - ab + ac - ab + b(1+1) - bc -ac + bc - c(1+1)

= a2 + (- ab - ab) + (ac -ac) + b2 + (- bc + bc) - c2

= a2 - 2ab + b2 - c2

Hence, option 2 is the correct option.

Question 2(i)

Use direct method to evaluate the following product:

(a - 8) (a + 2)

Answer

(a - 8) (a + 2)

Using the formula,

(x - a)(x + b) = x2 - (a - b)x - ab

= a2 - (8 - 2)a - 8 ×\times 2

= a2 - 6a - 16

Hence, (a - 8) (a + 2) = a2 - 6a - 16

Question 2(ii)

Use direct method to evaluate the following product:

(b - 3) (b - 5)

Answer

(b - 3) (b - 5)

Using the formula,

(x + a)(x + b) = x2 - (a + b)x + ab

= b2 - (3 + 5)b + 3 ×\times 5

= b2 - 8b + 15

Hence, (b - 3) (b - 5) = b2 - 8b + 15

Question 2(iii)

Use direct method to evaluate the following product:

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

Answer

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

= (3x ×\times 2x) + (3x ×\times y + (-2y) ×\times 2x) + ((-2y) ×\times y)

= 6x2 + (3xy - 4xy) - 2y2

= 6x2 - 1xy - 2y2

Hence, (3x - 2y) (2x + y) = 6x2 - xy - 2y2

Question 2(iv)

Use direct method to evaluate the following product:

(5a + 16) (3a - 7)

Answer

(5a + 16) (3a - 7)

= (5a ×\times 3a) + (5a ×\times (- 7) + 3a ×\times 16) + (16 ×\times (- 7))

= 15a2 + (-35a + 48a) - 112

= 15a2 + 13a - 112

Hence, (5a + 16) (3a - 7) = 15a2 + 13a - 112

Question 2(v)

Use direct method to evaluate the following product:

(8 - b) (3 + b)

Answer

(8 - b) (3 + b)

= (8 ×\times 3) + (8 ×\times b + (- b) ×\times 3) + ((- b) ×\times b)

= 24 + (8b - 3b) - b2

= 24 + 5b - b2

Hence, (8 - b) (3 + b) = 24 + 5b - b2

Question 3(i)

Evaluate:

(2a + 3) (2a - 3)

Answer

(2a + 3) (2a - 3)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (2a)2 - 32

= 4a2 - 9

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

Question 3(ii)

Evaluate:

(xy + 4) (xy - 4)

Answer

(xy + 4) (xy - 4)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (xy)2 - 42

= x2y2 - 16

Hence, (xy + 4) (xy - 4) = x2y2 - 16

Question 3(iii)

Evaluate:

(ab + x2) (ab - x2)

Answer

(ab + x2) (ab - x2)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (ab)2 - (x2)2

= a2b2 - x4

Hence, (ab + x2) (ab - x2) = a2b2 - x4

Question 3(iv)

Evaluate:

(3x2 + 5y2) (3x2 - 5y2)

Answer

(3x2 + 5y2) (3x2 - 5y2)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (3x2)2 - (5y2)2

= 9x4 - 25y4

Hence, (3x2 + 5y2) (3x2 - 5y2) = 9x4 - 25y4

Question 3(v)

Evaluate:

(z−23)\Big(z -\dfrac{2}{3}\Big) (z+23)\Big(z +\dfrac{2}{3}\Big)

Answer

(z−23)\Big(z -\dfrac{2}{3}\Big) (z+23)\Big(z +\dfrac{2}{3}\Big)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= z2 - (23)\Big(\dfrac{2}{3}\Big)2

= z2 - 49\dfrac{4}{9}

Hence, (z−23)\Big(z -\dfrac{2}{3}\Big) (z+23)\Big(z +\dfrac{2}{3}\Big) = z2 - 49\dfrac{4}{9}

Question 3(vi)

Evaluate:

(35a+12)\Big(\dfrac{3}{5}a +\dfrac{1}{2}\Big) (35a−12)\Big(\dfrac{3}{5}a -\dfrac{1}{2}\Big)

Answer

(35a+12)\Big(\dfrac{3}{5}a +\dfrac{1}{2}\Big) (35a−12)\Big(\dfrac{3}{5}a -\dfrac{1}{2}\Big)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (35a)2−(12)2\Big(\dfrac{3}{5}a\Big)^2 - \Big(\dfrac{1}{2}\Big)^2

= (925a2)−14\Big(\dfrac{9}{25}a^2\Big) - \dfrac{1}{4}

Hence, (35a+12)\Big(\dfrac{3}{5}a +\dfrac{1}{2}\Big) (35a−12)\Big(\dfrac{3}{5}a -\dfrac{1}{2}\Big) = (925a2)−14\Big(\dfrac{9}{25}a^2\Big) - \dfrac{1}{4}

Question 3(vii)

Evaluate:

(0.5 - 2a) (0.5 + 2a)

Answer

(0.5 - 2a) (0.5 + 2a)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (0.5)2 - (2a)2

= 0.25 - 4a2

Hence, (0.5 - 2a) (0.5 + 2a) = 0.25 - 4a2

Question 3(viii)

Evaluate:

(a2−b3)\Big(\dfrac{a}{2} -\dfrac{b}{3}\Big) (a2+b3)\Big(\dfrac{a}{2} +\dfrac{b}{3}\Big)

Answer

(a2−b3)\Big(\dfrac{a}{2} -\dfrac{b}{3}\Big) (a2+b3)\Big(\dfrac{a}{2} +\dfrac{b}{3}\Big)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (a2)2−(b3)2\Big(\dfrac{a}{2}\Big)^2 - \Big(\dfrac{b}{3}\Big)^2

= (a24)−(b29)\Big(\dfrac{a^2}{4}\Big) - \Big(\dfrac{b^2}{9}\Big)

Hence, (a2−b3)\Big(\dfrac{a}{2} -\dfrac{b}{3}\Big) (a2+b3)\Big(\dfrac{a}{2} +\dfrac{b}{3}\Big) = (a24)−(b29)\Big(\dfrac{a^2}{4}\Big) - \Big(\dfrac{b^2}{9}\Big)

Question 4(i)

Evaluate:

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

Answer

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

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (a2 - b2) (a2 + b2)

= a4 - b4

Hence, (a + b) (a - b) (a2 + b2) = a4 - b4

Question 4(ii)

Evaluate:

(3 - 2x) (3 + 2x) (9 + 4x2)

Answer

(3 - 2x) (3 + 2x) (9 + 4x2)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= (32 - (2x)2) (9 + 4x2)

= (9 - 4x2) (9 + 4x2)

= 92 - (4x2)2

= 81 - 16x4

Hence, (3 - 2x) (3 + 2x) (9 + 4x2) = 81 - 16x4

Question 4(iii)

Evaluate:

(3x - 4y) (3x + 4y) (9x2 + 16y2)

Answer

(3x - 4y) (3x + 4y) (9x2 + 16y2)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= [(3x)2 - (4y)2] (9x2 + 16y2)

= (9x2 - 16y2) (9x2 + 16y2)

= (9x2)2 - (16y2)2

= 81x4 - 256y4

Hence, (3x - 4y) (3x + 4y) (9x2 + 16y2) = 81x4 - 256y4

Question 5

Use the formula: (a + b) (a - b) = a2 - b2 to evaluate:

(i) 21 x 19

(ii) 33 x 27

(iii) 103 x 97

(iv) 9.8 x 10.2

(v) 7.7 x 8.3

Answer

(i) 21 x 19

(20 + 1)(20 - 1)

Using the formula: (a + b) (a - b) = a2 - b2

⇒ (20 + 1)(20 - 1) = 202 - 12

= 400 - 1

= 399

Hence, 21 x 19 = 399.

(ii) 33 x 27

(30 + 3)(30 - 3)

Using the formula: (a + b) (a - b) = a2 - b2

⇒ (30 + 3)(30 - 3) = 302 - 32

= 900 - 9

= 891

Hence, 33 x 27 = 891.

(iii) 103 x 97

(100 + 3)(100 - 3)

Using the formula: (a + b) (a - b) = a2 - b2

⇒ (100 + 3)(100 - 3) = 1002 - 32

= 10000 - 9

= 9991

Hence, 103 x 97 = 9991.

(iv) 9.8 x 10.2

(10 - 0.2)(10 + 0.2)

Using the formula: (a + b) (a - b) = a2 - b2

⇒ (10 - 0.2)(10 + 0.2) = 102 - (0.2)2

= 100 - 0.04

= 99.96

Hence, 9.8 x 10.2 = 99.96.

(v) 7.7 x 8.3

(8 - 0.3)(8 + 0.3)

Using the formula: (a + b) (a - b) = a2 - b2

⇒ (8 - 0.3)(8 + 0.3) = 82 - (0.3)2

= 64 - 0.09

= 63.91

Hence, 7.7 x 8.3 = 63.91.

Question 6(i)

Evaluate:

(6 - xy) (6 + xy)

Answer

(6 - xy) (6 + xy)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= 62 - (xy)2

= 36 - x2y2

Hence, (6 - xy) (6 + xy) = 36 - x2y2

Question 6(ii)

Evaluate:

(7x+23y)\Big(7x +\dfrac{2}{3}y\Big) (7x−23y)\Big(7x -\dfrac{2}{3}y\Big)

Answer

(7x+23y)\Big(7x +\dfrac{2}{3}y\Big) (7x−23y)\Big(7x -\dfrac{2}{3}y\Big)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

=(7x)2+(23y)2=49x2+49y2= \Big(7x\Big)^2 +\Big(\dfrac{2}{3}y\Big)^2\\[1em] = 49x^2 + \dfrac{4}{9}y^2

Hence, (7x+23y)\Big(7x +\dfrac{2}{3}y\Big) (7x−23y)\Big(7x -\dfrac{2}{3}y\Big) = 49x2−(49y2)49x^2 - \Big(\dfrac{4}{9}y^2\Big)

Question 6(iii)

Evaluate:

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

Answer

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

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

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

= (a24b2)−(4b2a2)\Big(\dfrac{a^2}{4b^2}\Big) - \Big(\dfrac{4b^2}{a^2}\Big)

Hence, (a2b+2ba)\Big(\dfrac{a}{2b} +\dfrac{2b}{a}\Big) (a2b−2ba)\Big(\dfrac{a}{2b} -\dfrac{2b}{a}\Big) = (a24b2)−(4b2a2)\Big(\dfrac{a^2}{4b^2}\Big) - \Big(\dfrac{4b^2}{a^2}\Big)

Question 6(iv)

Evaluate:

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

Answer

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

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

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

Hence, (3x−12y)\Big(3x -\dfrac{1}{2y}\Big) (3x+12y)\Big(3x +\dfrac{1}{2y}\Big) = 9x2−(14y2)9x^2 - \Big(\dfrac{1}{4y^2}\Big)

Question 6(v)

Evaluate:

(2a + 3) (2a - 3) (4a2 + 9)

Answer

(2a + 3) (2a - 3) (4a2 + 9)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= [(2a)2 - 32] (4a2 + 9)

= (4a2 - 92) (4a2 + 9)

= (4a2)2 - 92

= 16a4 - 81

Hence, (2a + 3) (2a - 3) (4a2 + 9) = 16a4 - 81

Question 6(vi)

Evaluate:

(a + bc) (a - bc) (a2 + b2c2)

Answer

(a + bc) (a - bc) (a2 + b2c2)

Using the formula

[∵ (x + y)(x - y) = x2 - y2]

= [a2 - (bc)2] (a2 + b2c2)

= (a2 - b2c2) (a2 + b2c2)

= (a2)2 - (b2c2)2

= a4 - b4c4

Hence, (a + bc) (a - bc) (a2 + b2c2) = a4 - b4c4

Question 7(i)

Expand:

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

Answer

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

Using the formula:

(∵ (x + y)2 = x2 + 2xy + y2)

=(a)2+2×a×12a+(12a)2=a2+2a2a+14a2=a2+1+14a2= (a)^2 + 2 \times a \times \dfrac{1}{2a} + \Big(\dfrac{1}{2a}\Big)^2\\[1em] = a^2 + \dfrac{2a}{2a} + \dfrac{1}{4a^2}\\[1em] = a^2 + 1 + \dfrac{1}{4a^2}

Hence, (a+12a)2\Big(a +\dfrac{1}{2a}\Big)^2 = a2+1+(14a2)a^2 + 1 + \Big(\dfrac{1}{4a^2}\Big)

Question 7(ii)

Expand:

(2a−1a)2\Big(2a -\dfrac{1}{a}\Big)^2

Answer

(2a−1a)2\Big(2a -\dfrac{1}{a}\Big)^2

Using the formula:

(∵ (x + y)2 = x2 + 2xy + y2)

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

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

Question 7(iii)

Expand:

(a + b - c)2

Answer

(a + b - c)2

Using the formula,

[∵ (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz]

= a2 + b2 + c2 + 2ab - 2bc - 2ca

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

Question 7(iv)

Expand:

(a - b + c)2

Answer

(a - b + c)2

Using the formula,

[∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz]

= a2 + b2 + c2 - 2ab - 2bc + 2ca

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

Question 7(v)

Expand:

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

Answer

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

Using the formula

(∵ (x + y)2 = x2 + 2xy + y2)

=(3x)2+2×3x×13x+13x2=9x2+6x3x+19x2=9x2+2+19x2= (3x)^2 + 2 \times 3x \times \dfrac{1}{3x} + \dfrac{1}{3x}^2\\[1em] = 9x^2 + \dfrac{6x}{3x} + \dfrac{1}{9x^2}\\[1em] = 9x^2 + 2 + \dfrac{1}{9x^2} Hence, (3x+13x)2=9x2+2+19x2\Big(3x +\dfrac{1}{3x}\Big)^2 = 9x^2 + 2 + \dfrac{1}{9x^2}

Question 8(i)

Find the square of:

a+15aa +\dfrac{1}{5a}

Answer

(a+15a)2(a +\dfrac{1}{5a})^2

Using the formula

(∵ (x + y)2 = x2 + 2xy + y2)

=a2+2×a×15a+15a2=a2+2a5a+125a2=a2+25+125a2= a^2 + 2 \times a \times \dfrac{1}{5a} + \dfrac{1}{5a}^2\\[1em] = a^2 + \dfrac{2a}{5a} + \dfrac{1}{25a^2}\\[1em] = a^2 + \dfrac{2}{5} + \dfrac{1}{25a^2}

Hence, (a+15a)2(a +\dfrac{1}{5a})^2 = a2 + 25\dfrac{2}{5} + 125a2\dfrac{1}{25a^2}

Question 8(ii)

Find the square of:

2a−1a2a -\dfrac{1}{a}

Answer

(2a−1a)2(2a - \dfrac{1}{a})^2

Using the formula

(∵ (x - y)2 = x2 - 2xy + y2)

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

Hence, (2a−1a)2(2a - \dfrac{1}{a})^2 = 4a2−4+1a24a^2 - 4 + \dfrac{1}{a^2}

Question 8(iii)

Find the square of:

x - 2y + 1

Answer

(x - 2y + 1)2

Using the formula

(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)

= x2 + (2y)2 + 12 - 2 ×\times (x) ×\times (2y) - 2 ×\times (2y) ×\times 1 + 2 ×\times 1 ×\times (x)

= x2 + 4y2 + 1 - 4xy - 4y + 2x

Hence,(x - 2y + 1)2 = x2 + 4y2 + 1 - 4xy - 4y + 2x

Question 8(iv)

Find the square of:

3a - 2b - 5c

Answer

(3a - 2b - 5c)2

Using the formula

(∵ (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz)

= (3a)2 + (2b)2 + (5c)2 - 2 ×\times (3a) ×\times (2b) - 2 ×\times (2b) ×\times (5c) + 2 ×\times (5c) ×\times (3a)

= 9a2 + 4b2 + 25c2 - 12ab + 20bc - 30ca

Hence, (3a - 2b - 5c)2 = 9a2 + 4b2 + 25c2 - 12ab + 20bc - 30ca

Question 8(v)

Find the square of:

2x+1x+12x +\dfrac{1}{x}+ 1

Answer

(2x+1x+1)(2x +\dfrac{1}{x}+ 1)2

Using the formula

(∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz)

=(2x)2+(1x)2+12+2×2x×(1x)+2×(1x)×1+2×1×2x=4x2+(1x2)+1+(4xx)+(2x)+4x=4x2+(1x2)+1+4+(2x)+4x=4x2+(1x2)+5+(2x)+4x= (2x)^2 + \Big(\dfrac{1}{x}\Big)^2 + 1^2 + 2 \times 2x \times \Big(\dfrac{1}{x}\Big) + 2 \times \Big(\dfrac{1}{x}\Big) \times 1 + 2 \times 1 \times 2x\\[1em] = 4x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + \Big(\dfrac{4x}{x}\Big) + \Big(\dfrac{2}{x}\Big) + 4x\\[1em] = 4x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + 4 + \Big(\dfrac{2}{x}\Big) + 4x\\[1em] = 4x^2 + \Big(\dfrac{1}{x^2}\Big) + 5 + \Big(\dfrac{2}{x}\Big) + 4x

Hence, (2x+1x+1)(2x +\dfrac{1}{x}+ 1)2 = 4x2 + (1x2)\Big(\dfrac{1}{x^2}\Big) + 5 + (2x)\Big(\dfrac{2}{x}\Big) + 4x

Question 8(vi)

Find the square of:

5−x+2x5 - x +\dfrac{2}{x}

Answer

(5−x+2x)2(5 - x +\dfrac{2}{x})^2

Using the formula

(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)

=(5)2+(x)2+(2x)2−2×5×x−2×x×(2x)+2×(2x)×5=25+x2+(4x2)−10x−(4xx)+(20x)=25+x2+(4x2)−10x−4+(20x)=(25−4)+x2+(4x2)−10x+(20x)=21+x2+(4x2)−10x+(20x)= (5)^2 + (x)^2 + \Big(\dfrac{2}{x}\Big)^2 - 2 \times 5 \times x - 2 \times x \times \Big(\dfrac{2}{x}\Big) + 2 \times \Big(\dfrac{2}{x}\Big) \times 5\\[1em] = 25 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x - \Big(\dfrac{4x}{x}\Big) + \Big(\dfrac{20}{x}\Big)\\[1em] = 25 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x - 4 + \Big(\dfrac{20}{x}\Big)\\[1em] = (25 - 4) + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x + \Big(\dfrac{20}{x}\Big)\\[1em] = 21 + x^2 + \Big(\dfrac{4}{x^2}\Big) - 10x + \Big(\dfrac{20}{x}\Big)

Hence, (5−x+2x)2(5 - x +\dfrac{2}{x})^2 = 21 + x2 + (4x2)\Big(\dfrac{4}{x^2}\Big) - 10x + (20x)\Big(\dfrac{20}{x}\Big)

Question 8(vii)

Find the square of:

2x - 3y + z

Answer

(2x - 3y + z)2

Using the formula

(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)

= (2x)2 + (3y)2 + (z)2 - 2 ×\times (2x) ×\times (3y) - 2 ×\times (3y) ×\times (z) + 2 ×\times (z) ×\times (2x)

= 4x2 + 9y2 + z2 - 12xy - 6yz + 4xz

Hence, (2x - 3y + z)2 = 4x2 + 9y2 + z2 - 12xy - 6yz + 4xz

Question 8(viii)

Find the square of:

x+1x−1x +\dfrac{1}{x}- 1

Answer

(x+1x−1)2(x +\dfrac{1}{x}- 1)^2

Using the formula

(∵ (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz)

=(x)2+(1x)2+(1)2+2×x×(1x)−2×(1x)×1−2×1×x=x2+(1x2)+1+(2xx)−(2x)−2x=x2+(1x2)+1+2−(2x)−2x=x2+(1x2)+3−(2x)−2x= (x)^2 + \Big(\dfrac{1}{x}\Big)^2 + (1)^2 + 2 \times x \times \Big(\dfrac{1}{x}\Big) - 2 \times \Big(\dfrac{1}{x}\Big) \times 1 - 2 \times 1 \times x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + \Big(\dfrac{2x}{x}\Big) - \Big(\dfrac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 1 + 2 - \Big(\dfrac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\dfrac{1}{x^2}\Big) + 3 - \Big(\dfrac{2}{x}\Big) - 2x

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

Question 9

Evaluate using expansion of (a + b)2 or (a - b)2 :

(i) (208)2

(ii) (92)2

(iii) (9.4)2

(iv) (20.7)2

Answer

(i) (208)2

(200 + 8)2

Using the formula

(∵ (x + y)2 = x2 + 2xy + y2)

= (200)2 + 2 x 200 x 8 + (8)2

= 40000 + 3200 + 64

= 43264

Hence, (208)2 = 43264

(ii) (92)2

(100 - 8)2

Using the formula

(∵ (x - y)2 = x2 - 2xy + y2)

= (100)2 - 2 x 100 x 8 + (8)2

= 10000 - 1600 + 64

= 8400 + 64

= 8464

Hence, (92)2 = 8464

(iii) (9.4)2

(10 - 0.6)2

Using the formula

(∵ (x - y)2 = x2 - 2xy + y2)

= (10)2 - 2 x 10 x 0.6 + (0.6)2

= 100 - 12 + 0.36

= 88 + 0.36

= 88.36

Hence, (9.4)2 = 88.36

(iv) (20.7)2

(20 + 0.7)2

Using the formula

(∵ (x + y)2 = x2 + 2xy + y2)

= (20)2 + 2 x 20 x 0.7 + (0.7)2

= 400 + 28 + 0.49

= 428 + 0.49

= 428.49

Hence, (20.7)2 = 428.49

Question 10(i)

Expand:

(2a + b)3

Answer

(2a + b)3

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

= (2a)3 + b3 + 3 x (2a)2x (b) + 3 x (2a) x (b)2

= 8a3 + b3 + 12a2b + 6ab2

Hence, (2a + b)3 = 8a3 + b3 + 12a2b + 6ab2

Question 10(ii)

Expand:

(a - 2b)3

Answer

(a - 2b)3

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

= a3 - (2b)3 - 3 x (a)2x (2b) + 3 x (a) x (2b)2

= a3 - 8b3 - 6a2b + 12ab2

Hence, (a - 2b)3 = a3 - 8b3 - 6a2b + 12ab2

Question 10(iii)

Expand:

(3x - 2y)3

Answer

(3x - 2y)3

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

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

= 27x3 - 8y3 - 54x2y + 36xy2

Hence, (3x - 2y)3 = 27x3 - 8y3 - 54x2y + 36xy2

Question 10(iv)

Expand:

(x + 5y)3

Answer

(x + 5y)3

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

= (x)3 + (5y)3 + 3 ×\times (x)2 ×\times (5y) + 3 ×\times (x) ×\times (5y)2

= x3 + 125y3 + 15x2y + 75xy2

Hence, (x + 5y)3 = x3 + 125y3 + 15x2y + 75xy2

Question 10(v)

Expand:

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

Answer

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

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

=(a)3+(1a)3+3×(a)2×(1a)+3×(a)×(1a)2=a3+(1a3)+(3a2a)+(3aa2)=a3+(1a3)+3a+(3a)= (a)^3 + \Big(\dfrac{1}{a}\Big)^3 + 3 \times (a)^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times (a) \times \Big(\dfrac{1}{a}\Big)^2\\[1em] = a^3 + \Big(\dfrac{1}{a^3}\Big) + \Big(\dfrac{3a^2}{a}\Big) + \Big(\dfrac{3a}{a^2}\Big)\\[1em] = a^3 + \Big(\dfrac{1}{a^3}\Big) + 3a + \Big(\dfrac{3}{a}\Big)\\[1em]

Hence, (a+1a)3=a3+(1a3)+3a+(3a)\Big(a +\dfrac{1}{a}\Big)^3 = a^3 + \Big(\dfrac{1}{a^3}\Big) + 3a + \Big(\dfrac{3}{a}\Big)

Question 10(vi)

Expand:

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

Answer

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

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

=(2a)3−(12a)3−3×(2a)2×(12a)+3×(2a)×(12a)2=8a3−(18a3)−(12a22a)+(6a4a2)=8a3−(18a3)−6a+(32a)= (2a)^3 - \Big(\dfrac{1}{2a}\Big)^3 - 3 \times (2a)^2 \times \Big(\dfrac{1}{2a}\Big) + 3 \times (2a) \times \Big(\dfrac{1}{2a}\Big)^2\\[1em] = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - \Big(\dfrac{12a^2}{2a}\Big) + \Big(\dfrac{6a}{4a^2}\Big)\\[1em] = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - 6a + \Big(\dfrac{3}{2a}\Big)\\[1em]

Hence, (2a−12a)3=8a3−(18a3)−6a+(32a)\Big(2a - \dfrac{1}{2a}\Big)^3 = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - 6a + \Big(\dfrac{3}{2a}\Big)

Question 11(i)

Find the cube of:

a + 2

Answer

(a + 2)3

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

= (a)3 + (2)3 + 3 ×\times (a)2 ×\times (2) + 3 ×\times (a) ×\times (2)2

= a3 + 8 + 6a2 + 12a

Hence, (a + 2)3 = a3 + 8 + 6a2 + 12a

Question 11(ii)

Find the cube of:

2a - 1

Answer

(2a - 1)3

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

= (2a)3 - (1)3 - 3 ×\times (2a)2 ×\times (1) + 3 ×\times (2a) ×\times (1)2

= 8a3 - 1 - 12a2 + 6a

Hence, (2a - 1)3 = 8a3 - 1 - 12a2 + 6a

Question 11(iii)

Find the cube of:

2a + 3b

Answer

(2a + 3b)3

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

= (2a)3 + (3b)3 + 3 ×\times (2a)2 ×\times (3b) + 3 ×\times (2a) ×\times (3b)2

= 8a3 + 27b3 + 36a2b + 54ab2

Hence, (2a + 3b)3 = 8a3 + 27b3 + 36a2b + 54ab2

Question 11(iv)

Find the cube of:

3b - 2a

Answer

(3b - 2a)3

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

= (3b)3 - (2a)3 + 3 ×\times (3b)2 ×\times (2a) + 3 ×\times (3b) ×\times (2a)2

= 27b3 - 8a3 - 54b2a + 36ba2

Hence, (3b - 2a)3 = 27b3 - 8a3 - 54b2a + 36ba2

Question 11(v)

Find the cube of:

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

Answer

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

Using the formula,

(x + y)3 = x3 + y3 + 3x2y + 3xy2

=(2x)3+(1x)3+3×(2x)2×(1x)+3×2x×(1x)2=8x3+(1x3)+(12x2x)+(6xx2)=8x3+(1x3)+12x+6x= (2x)^3 + \Big(\dfrac{1}{x}\Big)^3 + 3 \times (2x)^2 \times \Big(\dfrac{1}{x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{x}\Big)^2\\[1em] = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + \Big(\dfrac{12x^2}{x}\Big) + \Big(\dfrac{6x}{x^2}\Big)\\[1em] = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + 12x + \dfrac{6}{x}\\[1em]

Hence, 2x+1x=8x3+(1x3)+12x+6x2x + \dfrac{1}{x} = 8x^3 + \Big(\dfrac{1}{x^3}\Big) + 12x + \dfrac{6}{x}

Question 11(vi)

Find the cube of:

x−12x -\dfrac{1}{2}

Answer

x−12x - \dfrac{1}{2}

Using the formula,

(x - y)3 = x3 - y3 - 3x2y + 3xy2

=(x)3−(12)3−3×x2×(12)+3×x×(12)2=x3−(18)−(3x22)+(3x4)= (x)^3 - \Big(\dfrac{1}{2}\Big)^3 - 3 \times x^2 \times \Big(\dfrac{1}{2}\Big) + 3 \times x \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = x^3 - \Big(\dfrac{1}{8}\Big) - \Big(\dfrac{3x^2}{2}\Big) + \Big(\dfrac{3x}{4}\Big)

Hence, x+12=x3−(18)−(3x22)+(3x4)x + \dfrac{1}{2} = x^3 - \Big(\dfrac{1}{8}\Big) - \Big(\dfrac{3x^2}{2}\Big) + \Big(\dfrac{3x}{4}\Big)

Exercise 12(B)

Question 1(i)

If x−1x=3x -\dfrac{1}{x}= 3, the value of x2+1x2x^2 +\dfrac{1}{x^2} is:

  1. 0

  2. 7

  3. 11

  4. 9

Answer

Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

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

Putting the value, x−1x=3x -\dfrac{1}{x}= 3

(3)2=x2−2+1x2⇒9=x2−2+1x2⇒x2+1x2=9+2⇒x2+1x2=11(3)^2 = x^2 - 2 + \dfrac{1}{x^2}\\[1em] ⇒ 9 = x^2 - 2 + \dfrac{1}{x^2}\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 9 + 2\\[1em] ⇒ x^2 + \dfrac{1}{x^2} = 11

Hence, option 3 is the correct option.

Question 1(ii)

If a + b = 7 and ab = 10; the value of a2 + b2 is equal to:

  1. 29

  2. 49

  3. 39

  4. 69

Answer

Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value, a + b = 7 and ab = 10

⇒ (7)2 = a2 + 2 x 10 + b2

⇒ 49 = a2 + 20 + b2

⇒ a2 + b2 = 49 - 20

⇒ a2 + b2 = 29

Hence, option 1 is the correct option.

Question 1(iii)

The value of 95×95−5×595−5\dfrac{95 \times 95 - 5 \times 5}{95 - 5} is equal to:

  1. 100

  2. 90

  3. 95

  4. none of these

Answer

Using the formula,

[∵(x2 - y2) = (x + y)(x + y)]

95×95−5×595−5(95)2−(5)295−5=(95−5)(95+5)95−5=(95−5)(95+5)(95−5)=95+5=100\dfrac{95 \times 95 - 5 \times 5}{95 - 5}\\[1em] \dfrac{(95)^2 - (5)^2}{95 - 5}\\[1em] = \dfrac{(95 - 5)(95 + 5)}{95 - 5}\\[1em] = \dfrac{\cancel{(95 - 5)}(95 + 5)}{\cancel{(95 - 5)}}\\[1em] = 95 + 5\\[1em] = 100

Hence, option 1 is the correct option.

Question 1(iv)

If a - b = 1 and a + b = 3, the value of ab is:

  1. 4

  2. 2

  3. -2

  4. 0

Answer

Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

And

[∵(x - y)2 = x2 - 2xy + y2]

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

Putting a + b = 3, we get

(3)2 = a2 + 2ab + b2

9 = a2 + 2ab + b2

9 - 2ab = a2 + b2 ...............(1)

And,

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

Putting the value, a - b = 1

(1)2 = a2 - 2ab + b2

1 = a2 + b2 - 2ab

Using equation (1)

1 = 9 - 2ab - 2ab

1 = 9 - 4ab

4ab = 9 - 1

4ab = 8

ab = 84\dfrac{8}{4}

ab = 2

Hence, option 2 is the correct option.

Question 1(v)

If x+1x=2x +\dfrac{1}{x} = 2, the value of (x3+1x3)−(x2+1x2)\Big(x^3 +\dfrac{1}{x^3}\Big) - \Big(x^2 +\dfrac{1}{x^2}\Big) is:

  1. 0

  2. 4

  3. 2

  4. 6

Answer

Using the formula,

[∵ (x + y)3 = x3 + 3xy(x + y) + y3]

And,

[∵ (x + y)2 = x2 + 2xy + y2]

(x+1x)3=x3+3×x×1x(x+1x)+(1x)3⇒(x+1x)3=x3+3(x+1x)+1x3\Big(x + \dfrac{1}{x}\Big)^3 = x^3 + 3 \times x \times \dfrac{1}{x}\Big(x + \dfrac{1}{x}\Big) + \Big(\dfrac{1}{x}\Big)^3\\[1em] ⇒ \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + 3\Big(x + \dfrac{1}{x}\Big) + \dfrac{1}{x^3}

Putting x+1x=2x +\dfrac{1}{x} = 2

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

And ,

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

Putting the value x+1x=2x +\dfrac{1}{x} = 2,we get

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

Now, using equation (1) and (2),

(x3+1x3)−(x2+1x2)=2−2=0\Big(x^3 +\dfrac{1}{x^3}\Big) - \Big(x^2 +\dfrac{1}{x^2}\Big)\\[1em] = 2 - 2\\[1em] = 0

Hence, option 1 is the correct option.

Question 2

If a + b = 5 and ab = 6, find a2 + b2

Answer

Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value, a + b = 5 and ab = 6

⇒ (5)2 = a2 + 2 x 6 + b2

⇒ 25 = a2 + 12 + b2

⇒ a2 + b2 = 25 - 12

⇒ a2 + b2 = 13

Hence, the value of (a2 + b2) is 13.

Question 3

If a - b = 6 and ab = 16, find a2 + b2

Answer

Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value, a - b = 6 and ab = 16

⇒ (6)2 = a2 - 2 x 16 + b2

⇒ 36 = a2 - 32 + b2

⇒ a2 + b2 = 36 + 32

⇒ a2 + b2 = 68

Hence, the value of (a2 + b2) is 68.

Question 4

If a2 + b2 = 29 and ab = 10, find:

(i) a + b

(ii) a - b

Answer

(i) Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value, a2 + b2 = 29 and ab = 10

⇒ (a + b)2 = (a2 + b2) + 2ab

⇒ (a + b)2 = (29) + 2 ×\times 10

⇒ (a + b)2 = 29 + 20

⇒ (a + b)2 = 49

⇒ a + b = 49\sqrt{49}

⇒ a + b = 7 or -7

Hence, the values of (a + b) are 7 or -7.

(ii) Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value, a2 + b2 = 29 and ab = 10

⇒ (a - b)2 = (a2 + b2) - 2ab

⇒ (a - b)2 = (29) - 2 ×\times 10

⇒ (a - b)2 = 29 - 20

⇒ (a - b)2 = 9

⇒ a - b = 9\sqrt{9}

⇒ a - b = 3 or -3

Hence, the values of (a - b) are 3 or -3.

Question 5

If a2 + b2 = 10 and ab = 3, find:

(i) a - b

(ii) a + b

Answer

(i) Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value, a2 + b2 = 10 and ab = 3

⇒ (a - b)2 = (a2 + b2) - 2ab

⇒ (a - b)2 = (10) - 2 ×\times 3

⇒ (a - b)2 = 10 - 6

⇒ (a - b)2 = 4

⇒ a - b = 4\sqrt{4}

⇒ a - b = 2 or -2

Hence, the values of (a - b) are 2 or -2.

(ii) Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value, a2 + b2 = 10 and ab = 3

⇒ (a + b)2 = (a2 + b2) + 2ab

⇒ (a + b)2 = (10) + 2 ×\times 3

⇒ (a + b)2 = 10 + 6

⇒ (a + b)2 = 16

⇒ a + b = 16\sqrt{16}

⇒ a + b = 4 or -4

Hence, the values of (a + b) are 4 or -4.

Question 6

If a+1a=3a +\dfrac{1}{a}= 3, find: a2+1a2a^2 +\dfrac{1}{a^2}

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(a+1a)2=a2+2×a×1a+(1a)2⇒(a+1a)2=a2+2+1a2\Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 + \dfrac{1}{a^2}

Putting the value a+1a=3a + \dfrac{1}{a} = 3,we get

32=a2+2+1a2⇒9=a2+2+1a2⇒a2+1a2=9−2⇒a2+1a2=73^2 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ 9 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 9 - 2 \\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 7

Hence, the value of a2+1a2a^2 + \dfrac{1}{a^2} is 7.

Question 7

If a−1a=4a -\dfrac{1}{a}= 4, find: a2+1a2a^2 +\dfrac{1}{a^2}

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value a−1a=4a - \dfrac{1}{a} = 4,we get

42=a2−2+1a2⇒16=a2−2+1a2⇒a2+1a2=16+2⇒a2+1a2=184^2 = a^2 - 2 + \dfrac{1}{a^2}\\[1em] ⇒ 16 = a^2 - 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 16 + 2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 18

Hence, the value of a2+1a2a^2 + \dfrac{1}{a^2} is 18.

Question 8

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

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(a+1a)2=a2+2×a×1a+(1a)2⇒(a+1a)2=a2+2+1a2\Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 + \dfrac{1}{a^2}

Putting the value a2+1a2=23a^2 +\dfrac{1}{a^2} = 23,we get

⇒(a+1a)2=(a2+1a2)+2⇒(a+1a)2=23+2⇒(a+1a)2=25⇒(a+1a)=25⇒(a+1a)=5 or−5⇒ \Big(a + \dfrac{1}{a}\Big)^2 = \Big(a^2 + \dfrac{1}{a^2}\Big) + 2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = 23 + 2\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = 25\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big) = \sqrt{25}\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big) = 5 \text{ or} -5

Hence, the values of (a+1a)\Big(a + \dfrac{1}{a}\Big) are 5 or -5.

Question 9

If a2+1a2=11a^2 +\dfrac{1}{a^2}= 11, find: a−1aa -\dfrac{1}{a}

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value a2+1a2=11a^2 + \dfrac{1}{a}^2 = 11,we get

⇒(a−1a)2=(a2+1a2)−2⇒(a−1a)2=11−2⇒(a−1a)2=9⇒(a−1a)=9⇒(a−1a)=3 or−3⇒ \Big(a - \dfrac{1}{a}\Big)^2 = \Big(a^2 + \dfrac{1}{a^2}\Big) - 2\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^2 = 11 - 2\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^2 = 9\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big) = \sqrt{9}\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big) = 3 \text{ or} -3

Hence, the values of (a−1a)\Big(a - \dfrac{1}{a}\Big) are 3 or -3.

Question 10

a + b + c = 10 and a2 + b2 + c2 = 38, find ab + bc + ca

Answer

Using the formula,

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

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

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

Putting (a + b + c) = 10 and (a2 + b2 + c2) = 38, we get

⇒ (10)2 = 38 + 2(ab + bc + ca)

⇒ 100 = 38 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 100 - 38

⇒ 2(ab + bc + ca) = 62

⇒ ab + bc + ca = 622\dfrac{62}{2}

⇒ ab + bc + ca = 31

Hence, the value of (ab + bc + ca) is 31.

Question 11

Find : a2 + b2 + c2, if a + b + c = 9 and ab + bc + ca = 24.

Answer

Using the formula,

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

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

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

Putting (a + b + c) = 9 and (ab + bc + ca) = 24, we get

⇒ (9)2 = a2 + b2 + c2 + 2 x 24

⇒ 81 = a2 + b2 + c2 + 48

⇒ a2 + b2 + c2 = 81 - 48

⇒ a2 + b2 + c2 = 33

Hence, the value of a2 + b2 + c2 is 33.

Question 12

Find: a + b + c, if a2 + b2 + c2 = 83 and ab + bc + ca = 71

Answer

Using the formula,

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

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

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

Putting (a2 + b2 + c2) = 83 and (ab + bc + ca) = 71, we get

⇒ (a + b + c)2 = 83 + 2 x 71

⇒ (a + b + c)2 = 83 + 142

⇒ (a + b + c)2 = 225

⇒ a + b + c = 225\sqrt{225}

⇒ a + b + c = 15 or - 15

Hence, the values of (a + b + c) are 15 or - 15.

Question 13

If a + b = 6 and ab = 8, find: a3 + b3.

Answer

Using the formula,

[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]

So,

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

Putting the values a + b = 6 and ab = 8, we get

⇒ (6)3 = a3 + b3 + 3 x 8 x 6

⇒ 216 = a3 + b3 + 144

⇒ a3 + b3 = 216 - 144

⇒ a3 + b3 = 72

Hence, the value of a3 + b3 is 72.

Question 14

If a - b = 3 and ab = 10, find: a3 - b3.

Answer

Using the formula,

[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]

So,

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

Putting the values a - b = 3 and ab = 10, we get

⇒ (3)3 = a3 - b3 - 3 x 10 x 3

⇒ 27 = a3 + b3 - 90

⇒ a3 + b3 = 27 + 90

⇒ a3 + b3 = 117

Hence, the value of a3 + b3 is 117.

Question 15

Find : a3+1a3a^3 +\dfrac{1}{a^3}, if a+1a=5a +\dfrac{1}{a}= 5.

Answer

Using the formula,

[∵ (x + y)3 = x3 + 3xy(x + y) + y3]

So,

(a+1a)3=a3+3×a×1a(a+1a)+(1a)3⇒(a+1a)3=a3+3(a+1a)+1a3\Big(a + \dfrac{1}{a}\Big)^3 = a^3 + 3 \times a \times \dfrac{1}{a}\Big(a + \dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^3\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + 3\Big(a + \dfrac{1}{a}\Big) + \dfrac{1}{a^3}

Putting a+1a=5a +\dfrac{1}{a} = 5

53=a3+3×5+1a3⇒125=a3+15+1a3⇒a3+1a3=125−15⇒a3+1a3=1105^3 = a^3 + 3 \times 5 + \dfrac{1}{a^3}\\[1em] ⇒ 125 = a^3 + 15 + \dfrac{1}{a^3}\\[1em] ⇒ a^3 + \dfrac{1}{a^3} = 125 - 15 \\[1em] ⇒ a^3 + \dfrac{1}{a^3} = 110

Hence, the value of a3+1a3a^3 + \dfrac{1}{a^3} is 110.

Question 16

Find : a3−1a3a^3 -\dfrac{1}{a^3}, if a−1a=4a -\dfrac{1}{a}= 4.

Answer

Using the formula,

[∵ (x - y)3 = x3 - 3xy(x - y) - y3]

So,

(a−1a)3=a3−3×a×1a(a−1a)−(1a)3⇒(a−1a)3=a3−3(a−1a)−1a3\Big(a - \dfrac{1}{a}\Big)^3 = a^3 - 3 \times a \times \dfrac{1}{a}\Big(a - \dfrac{1}{a}\Big) - \Big(\dfrac{1}{a}\Big)^3\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - 3\Big(a - \dfrac{1}{a}\Big) - \dfrac{1}{a^3}

Putting a−1a=4a - \dfrac{1}{a} = 4

43=a3−3×4−1a3⇒64=a3−12−1a3⇒a3−1a3=64+12⇒a3−1a3=764^3 = a^3 - 3 \times 4 - \dfrac{1}{a^3}\\[1em] ⇒ 64 = a^3 - 12 - \dfrac{1}{a^3}\\[1em] ⇒ a^3 - \dfrac{1}{a^3} = 64 + 12\\[1em] ⇒ a^3 - \dfrac{1}{a^3} = 76

Hence, the value of a3−1a3a^3 - \dfrac{1}{a^3} is 76.

Question 17

If 2x−12x=42x -\dfrac{1}{2x} = 4, find:

(i) 4x2+14x24x^2 +\dfrac{1}{4x^2}

(ii) 8x3−18x38x^3 -\dfrac{1}{8x^3}

Answer

(i) Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value 2x−12x=42x - \dfrac{1}{2x} = 4,we get

42=4x2−2+14x2⇒16=4x2−2+14x2⇒4x2+14x2=16+2⇒4x2+14x2=184^2 = 4x^2 - 2 + \dfrac{1}{4x^2}\\[1em] ⇒ 16 = 4x^2 - 2 + \dfrac{1}{4x^2}\\[1em] ⇒ 4x^2 + \dfrac{1}{4x^2} = 16 + 2 \\[1em] ⇒ 4x^2 + \dfrac{1}{4x^2} = 18

Hence, the value of 4x2+14x24x^2 + \dfrac{1}{4x^2} is 18.

(ii) Using the formula,

[∵ (x - y)3 = x3 - 3xy(x - y) - y3]

So,

(2x−12x)3=(2x)3−3×2x×12x(2x−12x)−(12x)3⇒(2x−12x)3=8x3−3(2x−12x)−18x3\Big(2x - \dfrac{1}{2x}\Big)^3 = (2x)^3 - 3 \times 2x \times \dfrac{1}{2x}\Big(2x - \dfrac{1}{2x}\Big) - \Big(\dfrac{1}{2x}\Big)^3\\[1em] ⇒ \Big(2x - \dfrac{1}{2x}\Big)^3 = 8x^3 - 3\Big(2x - \dfrac{1}{2x}\Big) - \dfrac{1}{8x^3}

Putting 2x−12x=42x - \dfrac{1}{2x} = 4

43=8x3−3×4−18x3⇒64=8x3−12−18x3⇒8x3−18x3=64+12⇒8x3−18x3=764^3 = 8x^3 - 3 \times 4 - \dfrac{1}{8x^3}\\[1em] ⇒ 64 = 8x^3 - 12 - \dfrac{1}{8x^3}\\[1em] ⇒ 8x^3 - \dfrac{1}{8x^3} = 64 + 12 \\[1em] ⇒ 8x^3 - \dfrac{1}{8x^3} = 76

Hence, the value of 8x3−18x38x^3 - \dfrac{1}{8x^3} = 76.

Question 18

If 3x+13x=33x +\dfrac{1}{3x} = 3, find:

(i) 9x2+19x29x^2 +\dfrac{1}{9x^2}

(ii) 27x3+127x327x^3 +\dfrac{1}{27x^3}

Answer

(i) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value 3x+13x=33x + \dfrac{1}{3x} = 3,we get

32=9x2+2+19x2⇒9=9x2+2+19x2⇒9x2+19x2=9−2⇒9x2+19x2=73^2 = 9x^2 + 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9 = 9x^2 + 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 9 - 2\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 7

Hence, the value of 9x2+19x29x^2 + \dfrac{1}{9x^2} is 7.

(ii) Using the formula,

[∵ (x + y)3 = x3 + 3xy(x + y) + y3]

So,

(3x+13x)3=(3x)3+3×3x×13x(3x+13x)+(13x)3⇒(3x+13x)3=27x3+3(3x+13x)+127x3\Big(3x + \dfrac{1}{3x}\Big)^3 = (3x)^3 + 3 \times 3x \times \dfrac{1}{3x}\Big(3x + \dfrac{1}{3x}\Big) + \Big(\dfrac{1}{3x}\Big)^3\\[1em] ⇒ \Big(3x + \dfrac{1}{3x}\Big)^3 = 27x^3 + 3\Big(3x + \dfrac{1}{3x}\Big) + \dfrac{1}{27x^3}

Putting 3x+13x=33x + \dfrac{1}{3x} = 3

33=27x3+3×3+127x3⇒27=27x3+9+127x3⇒27x3+127x3=27−9⇒27x3+127x3=183^3 = 27x^3 + 3 \times 3 + \dfrac{1}{27x^3}\\[1em] ⇒ 27 = 27x^3 + 9 + \dfrac{1}{27x^3}\\[1em] ⇒ 27x^3 + \dfrac{1}{27x^3} = 27 - 9 \\[1em] ⇒ 27x^3 + \dfrac{1}{27x^3} = 18

Hence, the value of 27x3+127x327x^3 + \dfrac{1}{27x^3} is 18.

Question 19

The sum of the squares of two numbers is 13 and their product is 6. Find :

(i) the sum of the two numbers.

(ii) the difference between them.

Answer

(i) Let 2 numbers be x and y. The sum of the squares of two numbers is 13 and their product is 6. So,

x2 + y2 = 13

And,

xy = 6

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

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

Putting the value,

⇒ (x + y)2 = 13 + 2 ×\times 6

⇒ (x + y)2 = 13 + 12

⇒ (x + y)2 = 25

⇒ x + y = 25\sqrt{25}

⇒ x + y = 5 or -5

Hence, the sum of the two numbers = 5 or -5.

(ii) Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

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

Putting the value,

⇒ (x - y)2 = 13 - 2 ×\times 6

⇒ (x - y)2 = 13 - 12

⇒ (x - y)2 = 1

⇒ x - y = 1\sqrt{1}

⇒ x - y = 1 or -1

Hence, the difference of the two numbers = 1 or -1.

Test Yourself

Question 1(i)

If a is a positive and a2+1a2=18a^2 +\dfrac{1}{a^2}= 18; then the value of a−1aa -\dfrac{1}{a} is:

  1. 4

  2. 16

  3. 20

  4. 2√5

Answer

Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

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

Putting the value, a2+1a2=18a^2 +\dfrac{1}{a^2}= 18

⇒(a−1a)2=a2+(1a)2−2⇒(a−1a)2=18−2⇒(a−1a)2=16⇒(a−1a)=16⇒(a−1a)=4 or−4⇒ \Big(a -\dfrac{1}{a}\Big)^2 = a^2 + \Big(\dfrac{1}{a}\Big)^2 - 2\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big)^2 = 18 - 2\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big)^2 = 16\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big) = \sqrt{16}\\[1em] ⇒ \Big(a -\dfrac{1}{a}\Big) = 4 \text{ or} -4

As a is a positive number.

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

Hence, option 1 is the correct option.

Question 1(ii)

(x + y)(x - y)(x2 + y2)(x4 + y4) is equal to:

  1. x4 + y4

  2. x8 + y8

  3. x8 - y8

  4. 2x6y6

Answer

Using the formula,

[∵(x - y)(x + y) = x2 - y2]

(x + y)(x - y)(x2 + y2)(x4 + y4)

⇒ [(x + y)(x - y)](x2 + y2)(x4 + y4)

⇒ [(x2 - y2)(x2 + y2)](x4 + y4)

⇒ [(x4 - y4)(x4 + y4)]

⇒ (x8 - y8)

Hence, option 3 is the correct option.

Question 1(iii)

The value of 102 x 98 is:

  1. 6999

  2. 6696

  3. 9696

  4. 9996

Answer

Using the formula,

[∵(x - y)(x + y) = x2 - y2]

102 x 98 = (100 + 2)(100 - 2)

= 1002 - 22

= 10000 - 4

= 9996

Hence, option 4 is the correct option.

Question 1(iv)

(x + 3) (x + 3) - (x - 2) (x - 2) is equal to:

  1. 10x + 5

  2. 10x - 5

  3. 5 - 10x

  4. none of these

Answer

Using the formula,

[∵(x + y) = x2 + y2 + 2xy]

And,

[∵(x - y) = x2 + y2 - 2xy]

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

⇒ [(x + 3)2] - [(x - 2)2]

⇒ [x2 + 32 + 2 ×\times x ×\times 3] - [x2 + 22 - 2 ×\times x ×\times 2]

⇒ [x2 + 9 + 6x] - [x2 + 4 - 4x]

⇒ x2 + 9 + 6x - x2 - 4 + 4x

⇒ (x2 - x2) + (9 - 4) + (6x + 4x)

⇒ 5 + 10x

⇒ 10x + 5

Hence, option 1 is the correct option.

Question 1(v)

If 5a = 302 - 252, the value of a is:

  1. 5

  2. 11

  3. 55

  4. none of these

Answer

Using the formula,

[∵(x - y)(x + y) = x2 - y2]

5a = 302 - 252

⇒ 5a = (30 - 25) ( 30 + 25)

⇒ 5a = 5 x 55

⇒ 5a = 275

⇒ a = 2755\dfrac{275}{5}

⇒ a = 55

Hence, option 3 is the correct option.

Question 1(vi)

Statement 1: Cube of a binomial : (a - b)3 = a3 + 3a2b - 3ab2 - b3

Statement 2: (a - b)2 - (a + b)2 = 4ab.

Which of the following options is correct?

  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

Cube of a binomial:

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

= a3 - b3 - 3a2b + 3ab2

= a3 - 3a2b + 3ab2 - b3

So, statement 1 is false.

Solving,

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

= [a2 - 2ab + b2] - [a2 + 2ab + b2]

= a2 - 2ab + b2 - a2 - 2ab - b2

= -4ab.

So, statement 2 is false.

Hence, Option 2 is the correct option.

Question 1(vii)

Assertion (A) : Use appropriate identity, we get 22.5 x 21.5 = 484.75.

Reason (R) : The product: (x + y)(x - y) = x2 - y2.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Solving,

(x + y)(x - y) = x(x - y) + y(x - y)

= x2 - xy + xy - y2

= x2 - y2

So, reason (R) is true.

Solving,

⇒ 22.5 x 21.5

⇒ (22 + 0.5)(22 - 0.5)

Using the identity; (x + y)(x - y) = x2 - y2

= (22)2 - (0.5)2

= 484 - 0.25

= 483.75

So, assertion (A) is false.

Hence, option 4 is the correct option.

Question 1(viii)

Assertion (A) : If we add 9 with 49x2 - 42x, the resultant expression will be a perfect square expression.

Reason (R) : The product of the sum and difference of the same two terms = Difference of their squares.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Given: 49x2 - 42x

Adding 9 in it the term the expression becomes

⇒ 49x2 - 42x + 9

⇒ (7x)2 - 2 × (7x) × 3 + (3)2

⇒ (7x - 3)2

The resultant expression will be a perfect square expression.

So, assertion (A) is true.

The product of the sum and difference of the same two terms = Difference of their squares.

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

So, reason (R) is true but reason (R) is not the correct explanation of assertion (A).

Hence, option 2 is the correct option.

Question 1(ix)

Assertion (A) : If the volume of a cube is a3 + b3 + 3ab(a + b), then the edge of the cube is (a + b)

Reason (R) : (1st term + 2nd term)3 = (1st term)3 + 3(1st term)2 .(2nd term) + 3(2nd term)2 .(1st term) + (2nd term)3.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

Volume of a cube = (edge)3

⇒ (edge)3 = a3 + b3 + 3ab(a + b)

⇒ (edge)3 = (a + b)3

⇒ edge = (a+b)33\sqrt[3]{(a + b)^3}

⇒ edge = (a + b)

So, assertion (A) is true.

Using identity,

⇒ (a + b)3 = a3 + b3 + 3ab(a + b)

⇒ (a + b)3 = a3 + 3a2b + 3ab2 + b3

Substitute a = 1st term and b = 2nd term

(1st term + 2nd term)3 = (1st term)3 + 3(1st term)2 .(2nd term) + 3(2nd term)2 .(1st term) + (2nd term)3

So, reason (R) is true and, reason (R) is the correct explanation of assertion (A).

Hence, option 1 is the correct option.

Question 1(x)

Assertion (A) : 687 x 687 - 313 x 313 = 37400

Reason (R) : The product of the sum and difference of the same two terms = The square of their difference.

  1. Both A and R are correct, and R is the correct explanation for A.

  2. Both A and R are correct, and R is not the correct explanation for A.

  3. A is true, but R is false.

  4. A is false, but R is true.

Answer

We know that,

The product of the sum and difference of the same two terms = Difference of their squares.

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

So, reason (R) is true.

Given,

⇒ 687 x 687 - 313 x 313 = 34700

Solving, L.H.S.

⇒ 687 x 687 - 313 x 313

⇒ 6872 - 3132

⇒ (687 - 313)(687 + 313)

⇒ 374 x 1000

⇒ 374000.

Since, L.H.S. ≠ R.H.S.

So, assertion (A) is false.

Hence, option 4 is the correct option.

Question 2(i)

Evaluate :

(3x+12)\Big(3x +\dfrac{1}{2}\Big) (2x+13)\Big(2x +\dfrac{1}{3}\Big)

Answer

(3x+12)(2x+13)=3x(2x+13)+12(2x+13)=3x×2x+3x×13+12×2x+12×13=6x2+3x3+2x2+1×12×3=6x2+x+x+16=6x2+2x+16\Big(3x +\dfrac{1}{2}\Big) \Big(2x +\dfrac{1}{3}\Big) \\[1em] = 3x \Big(2x +\dfrac{1}{3}\Big) + \dfrac{1}{2} \Big(2x +\dfrac{1}{3}\Big) \\[1em] = 3x \times 2x + 3x \times \dfrac{1}{3} + \dfrac{1}{2} \times 2x + \dfrac{1}{2} \times \dfrac{1}{3} \\[1em] = 6x^2 + \dfrac{3x}{3} + \dfrac{2x}{2} + \dfrac{1 \times 1}{2 \times 3} \\[1em] = 6x^2 + x + x + \dfrac{1}{6} \\[1em] = 6x^2 + 2x + \dfrac{1}{6}

Hence, (3x+12)(2x+13)=6x2+2x+16\Big(3x +\dfrac{1}{2}\Big) \Big(2x +\dfrac{1}{3}\Big) = 6x^2 + 2x + \dfrac{1}{6}

Question 2(ii)

Evaluate :

(2a + 0.5)(7a - 0.3)

Answer

(2a + 0.5)(7a - 0.3)

= 2a (7a - 0.3) + 0.5 (7a - 0.3)

= 14a(1+1) - 0.6a + 3.5a - 0.15

= 14a2 + 2.9a - 0.15

Hence, (2a + 0.5)(7a - 0.3) = 14a2 + 2.9a - 0.15

Question 2(iii)

Evaluate :

(9 - y) (7 + y)

Answer

(9 - y) (7 + y)

= 9 (7 + y) - y (7 + y)

= 63 + 9y - 7y - y(1+1)

= 63 + 2y - y2

Hence,(9 - y) (7 + y) = 63 + 2y - y2

Question 2(iv)

Evaluate :

(2 - z) (15 - z)

Answer

(2 - z) (15 - z)

= 2 (15 - z) - z (15 - z)

= 30 - 2z - 15z + z(1+1)

= 30 - 17z + z2

Hence, (2 - z) (15 - z) = 30 - 17z + z2

Question 2(v)

Evaluate :

(a2 + 5) (a2 - 3)

Answer

(a2 + 5) (a2 - 3)

= a2 (a2 - 3) + 5 (a2 - 3)

= a(2+2) - 3a2 + 5a2 - 15

= a4 + 2a2 - 15

Hence, (a2 + 5) (a2 - 3) = a4 + 2a2 - 15

Question 2(vi)

Evaluate :

(4 - ab) (8 + ab)

Answer

(4 - ab) (8 + ab)

= 4 (8 + ab) - ab (8 + ab)

= 32 + 4ab - 8ab - a(1+1)b(1+1)

= 32 - 4ab - a2b2

Hence, (4 - ab) (8 + ab) = 32 - 4ab - a2b2

Question 2(vii)

Evaluate :

(5xy - 7) (7xy + 9)

Answer

(5xy - 7) (7xy + 9)

= 5xy (7xy + 9) - 7 (7xy + 9)

= 35x(1+1)y(1+1) + 45xy - 49xy - 63

= 35x2y2 - 4xy - 63

Hence, (5xy - 7) (7xy + 9) = 35x2y2 - 4xy - 63

Question 2(viii)

Evaluate :

(3a2 - 4b2) (8a2 - 3b2)

Answer

(3a2 - 4b2) (8a2 - 3b2)

= 3a2 (8a2 - 3b2) - 4b2 (8a2 - 3b2)

= 24a(2+2) - 9a2b2 - 32a2b2 + 12b(2+2)

= 24a4 - 41a2b2 + 12b4

Hence, (3a2 - 4b2) (8a2 - 3b2) = 24a4 - 41a2b2 + 12b4

Question 3(i)

Find the square of:

3x+2y3x + \dfrac{2}{y}

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

(3x+2y)2=(3x)2+2×3x×2y+(2y)2=9x2+2×6xy+4y2=9x2+12xy+4y2\Big(3x +\dfrac{2}{y}\Big)^2\\[1em] = (3x)^2 + 2 \times 3x \times \dfrac{2}{y} + \Big(\dfrac{2}{y}\Big)^2\\[1em] = 9x^2 + \dfrac{2 \times 6x}{y} + \dfrac{4}{y^2}\\[1em] = 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}

Hence, (3x+2y)2=9x2+12xy+4y2\Big(3x +\dfrac{2}{y}\Big)^2= 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}

Question 3(ii)

Find the square of:

5a6b−6b5a\dfrac{5a}{6b} -\dfrac{6b}{5a}

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

(5a6b−6b5a)2=(5a6b)2−2×5a6b×6b5a+(6b5a)2=25a236b2−2×5a×6b6b×5a+36b225a2=25a236b2−60ab30ab+36b225a2=25a236b2−2+36b225a2\Big(\dfrac{5a}{6b} - \dfrac{6b}{5a}\Big)^2\\[1em] = \Big(\dfrac{5a}{6b}\Big)^2 - 2 \times \dfrac{5a}{6b} \times \dfrac{6b}{5a} + \Big(\dfrac{6b}{5a}\Big)^2\\[1em] = \dfrac{25a^2}{36b^2} - \dfrac{2 \times 5a \times 6b}{6b \times 5a} + \dfrac{36b^2}{25a^2}\\[1em] = \dfrac{25a^2}{36b^2} - \dfrac{60ab}{30ab} + \dfrac{36b^2}{25a^2}\\[1em] = \dfrac{25a^2}{36b^2} - 2 + \dfrac{36b^2}{25a^2}

Hence, (5a6b−6b5a)2=25a236b2−2+36b225a2\Big(\dfrac{5a}{6b} - \dfrac{6b}{5a}\Big)^2 = \dfrac{25a^2}{36b^2} - 2 + \dfrac{36b^2}{25a^2}

Question 3(iii)

Find the square of:

2m2−23n22m^2 -\dfrac{2}{3}n^2

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

(2m2−23n2)2=(2m2)2−2×2m2×23n2+(23n2)2=4m4−2×2m2×2n23+49n4=4m4−83m2n2+49n4\Big(2m^2 -\dfrac{2}{3}n^2\Big)^2\\[1em] = (2m^2)^2 - 2 \times 2m^2 \times \dfrac{2}{3}n^2 + \Big(\dfrac{2}{3}n^2\Big)^2\\[1em] = 4m^4 - \dfrac{2 \times 2m^2 \times 2n^2}{3} + \dfrac{4}{9}n^4\\[1em] = 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4

Hence, (2m2−23n2)2=4m4−83m2n2+49n4\Big(2m^2 -\dfrac{2}{3}n^2\Big)^2= 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4

Question 3(iv)

Find the square of:

5x+15x5x +\dfrac{1}{5x}

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

(5x+15x)2=(5x)2+2×5x×15x+(15x)2=25x2+10x5x+125x2=25x2+2+125x2\Big(5x +\dfrac{1}{5x}\Big)^2\\[1em] = (5x)^2 + 2 \times 5x \times \dfrac{1}{5x} + \Big(\dfrac{1}{5x}\Big)^2\\[1em] = 25x^2 + \dfrac{10x}{5x} + \dfrac{1}{25x^2}\\[1em] = 25x^2 + 2 + \dfrac{1}{25x^2}

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

Question 3(v)

Find the square of:

8x+32y8x +\dfrac{3}{2}y

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

(8x+32y)2=(8x)2+2×8x×32y+(32y)2=64x2+48xy2+94y2=64x2+24xy+94y2\Big(8x +\dfrac{3}{2}y\Big)^2\\[1em] = (8x)^2 + 2 \times 8x \times \dfrac{3}{2}y + \Big(\dfrac{3}{2}y\Big)^2\\[1em] = 64x^2 + \dfrac{48xy}{2} + \dfrac{9}{4}y^2\\[1em] = 64x^2 + 24xy + \dfrac{9}{4}y^2

Hence, (8x+32y)2=64x2+24xy+94y2\Big(8x +\dfrac{3}{2}y\Big)^2= 64x^2 + 24xy + \dfrac{9}{4}y^2

Question 3(vi)

Find the square of:

607

Answer

Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

6072

= (600 + 7)2

= (600)2 + 2 x 600 x 7 + (7)2

= 3,60,000 + 8,400 + 49

= 3,68,400 + 49

= 3,68,449

Hence, 6072 = 3,68,449

Question 3(vii)

Find the square of:

391

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

3912

= (400 - 9)2

= (400)2 - 2 x 400 x 9 + (9)2

= 1,60,000 - 7,200 + 81

= 1,52,800 + 81

=1,52,881

Hence, 3912 = 1,52,881

Question 3(viii)

Find the square of:

9.7

Answer

Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

9.72

= (10 - 0.3)2

= (10)2 - 2 x 10 x 0.3 + (0.3)2

= 100 - 6 + 0.09

= 94 + 0.09

= 94.09

Hence, 9.72 = 94.09

Question 4

If a+1a=2a +\dfrac{1}{a}= 2, find :

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

(ii) a4+1a4a^4 +\dfrac{1}{a^4}

Answer

(i) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(a+1a)2=a2+2×a×1a+1a2⇒(a+1a)2=a2+2+1a2\Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 \times a \times \dfrac{1}{a} + \dfrac{1}{a^2}\\[1em] ⇒ \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + 2 + \dfrac{1}{a^2}

Putting the value a+1a=2a + \dfrac{1}{a} = 2,we get

22=a2+2+1a2⇒4=a2+2+1a2⇒a2+1a2=4−2⇒a2+1a2=22^2 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ 4 = a^2 + 2 + \dfrac{1}{a^2}\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 4 - 2 \\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 2

Hence, the value of a2+1a2a^2 + \dfrac{1}{a^2} = 2.

(ii) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(a2+1a2)2=(a2)2+2×a2×1a2+(1a2)2⇒(a2+1a2)2=a4+2+1a4\Big(a^2 + \dfrac{1}{a^2}\Big)^2 = (a^2)^2 + 2 \times a^2 \times \dfrac{1}{a^2} + \Big(\dfrac{1}{a^2}\Big)^2\\[1em] ⇒ \Big(a^2 + \dfrac{1}{a^2}\Big)^2 = a^4 + 2 + \dfrac{1}{a^4}

Putting the value a2+1a2=2a^2 + \dfrac{1}{a^2} = 2,we get

22=a4+2+1a4⇒4=a4+2+1a4⇒a4+1a4=4−2⇒a4+1a4=22^2 = a^4 + 2 + \dfrac{1}{a^4}\\[1em] ⇒ 4 = a^4 + 2 + \dfrac{1}{a^4}\\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 4 - 2 \\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 2

Hence, the value of a4+1a4a^4 + \dfrac{1}{a^4} = 2.

Question 5

If m−1m=5m -\dfrac{1}{m}= 5, find:

(i) m2+1m2m^2 +\dfrac{1}{m^2}

(ii) m4+1m4m^4 +\dfrac{1}{m^4}

(iii) m2−1m2m^2 -\dfrac{1}{m^2}

Answer

(i) Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

(m−1m)2=m2−2×m×1m+1m2⇒(m−1m)2=m2−2+1m2\Big(m - \dfrac{1}{m}\Big)^2 = m^2 - 2 \times m \times \dfrac{1}{m} + \dfrac{1}{m^2}\\[1em] ⇒ \Big(m - \dfrac{1}{m}\Big)^2 = m^2 - 2 + \dfrac{1}{m^2}

Putting the value m−1m=5m - \dfrac{1}{m} = 5,we get

52=m2−2+1m2⇒25=m2−2+1m2⇒m2+1m2=25+2⇒m2+1m2=275^2 = m^2 - 2 + \dfrac{1}{m^2}\\[1em] ⇒ 25 = m^2 - 2 + \dfrac{1}{m^2}\\[1em] ⇒ m^2 + \dfrac{1}{m^2} = 25 + 2 \\[1em] ⇒ m^2 + \dfrac{1}{m^2} = 27

Hence, the value of m2+1m2m^2 + \dfrac{1}{m^2} = 27.

(ii) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(m2+1m2)2=(m2)2+2×m2×1m2+(1m2)2⇒(m2+1m2)2=m4+2+1m4\Big(m^2 + \dfrac{1}{m^2}\Big)^2 = (m^2)^2 + 2 \times m^2 \times \dfrac{1}{m^2} + \Big(\dfrac{1}{m^2}\Big)^2\\[1em] ⇒ \Big(m^2 + \dfrac{1}{m^2}\Big)^2 = m^4 + 2 + \dfrac{1}{m^4}

Putting the value m2+1m2=27m^2 + \dfrac{1}{m^2} = 27,we get

272=m4+2+1m4⇒729=m4+2+1m4⇒m4+1m4=729−2⇒m4+1m4=72727^2 = m^4 + 2 + \dfrac{1}{m^4}\\[1em] ⇒ 729 = m^4 + 2 + \dfrac{1}{m^4}\\[1em] ⇒ m^4 + \dfrac{1}{m^4} = 729 - 2 \\[1em] ⇒ m^4 + \dfrac{1}{m^4} = 727

Hence, the value of m4+1m4m^4 + \dfrac{1}{m^4} = 727.

(iii) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(m+1m)2=m2+2×m×1m+1m2⇒(m+1m)2=m2+2+1m2⇒(m+1m)2=m2+1m2+2\Big(m + \dfrac{1}{m}\Big)^2 = m^2 + 2 \times m \times \dfrac{1}{m} + \dfrac{1}{m}^2\\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = m^2 + 2 + \dfrac{1}{m^2}\\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = m^2 + \dfrac{1}{m^2} + 2

Putting the value m2+1m2=27m^2 + \dfrac{1}{m^2} = 27, we get

⇒(m+1m)2=27+2⇒(m+1m)2=29⇒(m+1m)=29⇒ \Big(m + \dfrac{1}{m}\Big)^2 = 27 + 2 \\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big)^2 = 29 \\[1em] ⇒ \Big(m + \dfrac{1}{m}\Big) = \sqrt{29} \\[1em]

Now, using the formula,

[∵ (x2 - y2) = (x - y)(x + y)]

So,

(m2−1m2)=(m−1m)(m+1m)\Big(m^2 - \dfrac{1}{m^2}\Big) = \Big(m - \dfrac{1}{m}\Big)\Big(m + \dfrac{1}{m}\Big)

Putting the value, (m−1m)=5\Big(m - \dfrac{1}{m}\Big) = 5 and (m+1m)=29\Big(m + \dfrac{1}{m}\Big) = \sqrt{29}

(m2−1m2)=(5)×(29)m2−1m2=529\Big(m^2 - \dfrac{1}{m^2}\Big) = (5) \times (\sqrt{29})\\[1em] m^2 - \dfrac{1}{m^2} = 5\sqrt{29}

Hence, the value of m2−1m2=529m^2 - \dfrac{1}{m^2} = 5\sqrt{29}.

Question 6

If a2 + b2 = 41 and ab = 4, find:

(i) a - b

(ii) a + b

Answer

(i) Using the formula,

[∵(x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value, a2 + b2 = 41 and ab = 4

⇒ (a - b)2 = (a2 + b2) - 2ab

⇒ (a - b)2 = (41) - 2 ×\times 4

⇒ (a - b)2 = 41 - 8

⇒ (a - b)2 = 33

⇒ a - b = 33\sqrt{33}

⇒ a - b = 33\sqrt{33}

Hence, the value of a - b = 33\sqrt{33}.

(ii) Using the formula,

[∵(x + y)2 = x2 + 2xy + y2]

So,

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

Putting the value, a2 + b2 = 41 and ab = 4

⇒ (a + b)2 = (a2 + b2) + 2ab

⇒ (a + b)2 = (41) + 2 ×\times 4

⇒ (a + b)2 = 41 + 8

⇒ (a + b)2 = 49

⇒ a + b = 49\sqrt{49}

⇒ a + b = 7

Hence, the value of a + b = 7.

Question 7

If 2a+12a=82a +\dfrac{1}{2a}= 8, find:

(i) 4a2+14a24a^2 +\dfrac{1}{4a^2}

(ii) 16a4+116a416a^4 +\dfrac{1}{16a^4}

Answer

(i) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(2a+12a)2=(2a)2+2×2a×12a+(12a)2⇒(2a+12a)2=4a2+2+14a2\Big(2a + \dfrac{1}{2a}\Big)^2 = (2a)^2 + 2 \times 2a \times \dfrac{1}{2a} + \Big(\dfrac{1}{2a}\Big)^2\\[1em] ⇒ \Big(2a + \dfrac{1}{2a}\Big)^2 = 4a^2 + 2 + \dfrac{1}{4a^2}

Putting the value 2a+12a=82a + \dfrac{1}{2a} = 8, we get

82=4a2+2+14a2⇒64=4a2+2+14a2⇒4a2+14a2=64−2⇒4a2+14a2=628^2 = 4a^2 + 2 + \dfrac{1}{4a^2}\\[1em] ⇒ 64 = 4a^2 + 2 + \dfrac{1}{4a^2}\\[1em] ⇒ 4a^2 + \dfrac{1}{4a^2} = 64 - 2 \\[1em] ⇒ 4a^2 + \dfrac{1}{4a^2} = 62

Hence, the value of 4a2+14a24a^2 + \dfrac{1}{4a^2} is 62.

(ii) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(4a2+14a2)2=(4a2)2+2×4a2×14a2+(14a2)2⇒(4a2+14a2)2=16a4+2+116a4\Big(4a^2 + \dfrac{1}{4a^2}\Big)^2 = (4a^2)^2 + 2 \times 4a^2 \times \dfrac{1}{4a^2} + \Big(\dfrac{1}{4a^2}\Big)^2\\[1em] ⇒ \Big(4a^2 + \dfrac{1}{4a^2}\Big)^2 = 16a^4 + 2 + \dfrac{1}{16a^4}

Putting the value 4a2+14a2=624a^2 + \dfrac{1}{4a^2} = 62, we get

622=16a4+2+116a4⇒3,844=16a4+2+116a4⇒16a4+116a4=3,844−2⇒16a4+116a4=3,84262^2 = 16a^4 + 2 + \dfrac{1}{16a^4}\\[1em] ⇒ 3,844 = 16a^4 + 2 + \dfrac{1}{16a^4}\\[1em] ⇒ 16a^4 + \dfrac{1}{16a^4} = 3,844 - 2 \\[1em] ⇒ 16a^4 + \dfrac{1}{16a^4} = 3,842

Hence, the value of 16a4+116a416a^4 + \dfrac{1}{16a^4} is 3,842.

Question 8

If 3x−13x=53x -\dfrac{1}{3x}= 5, find:

(i) 9x2+19x29x^2 +\dfrac{1}{9x^2}

(ii) 81x4+181x481x^4 +\dfrac{1}{81x^4}

Answer

(i) Using the formula,

[∵ (x - y)2 = x2 - 2xy + y2]

So,

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

Putting the value 3x−13x=53x - \dfrac{1}{3x} = 5,we get

52=9x2−2+19x2⇒25=9x2−2+19x2⇒9x2+19x2=25+2⇒9x2+19x2=275^2 = 9x^2 - 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 25 = 9x^2 - 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 25 + 2 \\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 27

Hence, the value of 9x2+19x29x^2 + \dfrac{1}{9x^2} is 27.

(ii) Using the formula,

[∵ (x + y)2 = x2 + 2xy + y2]

So,

(9x2+19x2)2=(9x2)2+2×9x2×19x2+(19x2)2⇒(9x2+19x2)2=81x4+2+181x4\Big(9x^2 + \dfrac{1}{9x^2}\Big)^2 = (9x^2)^2 + 2 \times 9x^2 \times \dfrac{1}{9x^2} + \Big(\dfrac{1}{9x^2}\Big)^2\\[1em] ⇒ \Big(9x^2 + \dfrac{1}{9x^2}\Big)^2 = 81x^4 + 2 + \dfrac{1}{81x^4}

Putting the value 9x2+19x2=279x^2 + \dfrac{1}{9x^2} = 27,we get

272=81x4+2+181x4⇒729=81x4+2+181x4⇒81x4+181x4=729−2⇒81x4+181x4=72727^2 = 81x^4 + 2 + \dfrac{1}{81x^4}\\[1em] ⇒ 729 = 81x^4 + 2 + \dfrac{1}{81x^4}\\[1em] ⇒ 81x^4 + \dfrac{1}{81x^4} = 729 - 2 \\[1em] ⇒ 81x^4 + \dfrac{1}{81x^4} = 727

Hence, the value of 81x4+181x481x^4 + \dfrac{1}{81x^4} is 727.

Question 9(i)

Expand:

(3x - 4y + 5z)2

Answer

Using the formula,

[∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz]

⇒ (3x - 4y + 5z)2 = (3x)2 + (4y)2 + (5z)2 - 2 ×\times (3x) ×\times (4y) - 2 ×\times (4y) ×\times (5z) + 2 ×\times (5z) ×\times (3x)

= 9x2 + 16y2 + 25z2 - 24xy - 40yz + 30xz

Hence, (3x - 4y + 5z)2 = 9x2 + 16y2 + 25z2 - 24xy - 40yz + 30xz.

Question 9(ii)

Expand:

(2a - 5b - 4c)2

Answer

Using the formula,

[∵ (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz]

⇒ (2a - 5b - 4c)2 = (2a)2 + (5b)2 + (4c)2 - 2 ×\times (2a) ×\times (5b) + 2 ×\times (5b) ×\times (4c) - 2 ×\times (4c) ×\times (2a)

= 4a2 + 25b2 + 16c2 - 20ab + 40bc - 16ca

Hence, (2a - 5b - 4c)2 = 4a2 + 25b2 + 16c2 - 20ab + 40bc - 16ca.

Question 9(iii)

Expand:

(5x + 3y)3

Answer

Using the formula,

[∵ (x + y)3 = x3 + y3 + 3x2y + 3xy2]

So,

⇒ (5x + 3y)3 = (5x)3 + (3y)3 + 3 ×\times (5x)2 ×\times (3y) + 3 ×\times(5x) ×\times (3y)2

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

Hence, (5x + 3y)3 = 125x3 + 27y3 + 225x2y + 135xy2.

Question 9(iv)

Expand:

(6a - 7b)3

Answer

Using the formula,

[∵ (x - y)3 = x3 - y3 - 3x2y + 3xy2]

So,

⇒ (6a - 7b)3 = (6a)3 - (7b)3 - 3 ×\times (6a)2 ×\times (7b) + 3 ×\times(6a) ×\times (7b)2

= 216a3 - 343b3 - 756a2b + 882ab2

Hence, (6a - 7b)3 = 216a3 - 343b3 - 756a2b + 882ab2.

Question 10

If a + b + c = 9 and ab + bc + ca = 15, find: a2 + b2 + c2.

Answer

Using the formula,

[∵(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

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

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

Putting the value a + b + c = 9 and ab + bc + ca = 15, we get

⇒ 92 = a2 + b2 + c2 + 2 x 15

⇒ 81 = a2 + b2 + c2 + 30

⇒ a2 + b2 + c2 = 81 - 30

⇒ a2 + b2 + c2 = 51

Hence, the value of a2 + b2 + c2 is 51.

Question 11

If a + b + c = 11 and a2 + b2 + c2 = 81, find: ab + bc + ca.

Answer

Using the formula,

[∵(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]

So,

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

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

Putting the value (a + b + c) = 11 and a2 + b2 + c2 = 81,

⇒ (11)2 = 81 + 2(ab + bc + ca)

⇒ 121 = 81 + 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 121 - 81

⇒ 2(ab + bc + ca) = 40

⇒ ab + bc + ca = 402\dfrac{40}{2}

⇒ ab + bc + ca = 20

Hence, the value of (ab + bc + ca) is 20.

Question 12

If 3x - 4y = 5 and xy = 3, find: 27x3 - 64y3

Answer

Using the formula,

[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]

So,

⇒ (3x - 4y)3 = (3x)3 - (4y)3 - 3 ×\times 3x ×\times 4y(3x - 4y)

⇒ (3x - 4y)3 = 27x3 - 64y3 - 36xy(3x - 4y)

Putting the value (3x - 4y) = 5 and xy = 3, we get

⇒ (5)3 = 27x3 - 64y3 - 36 ×\times 3 ×\times 5

⇒ 125 = 27x3 - 64y3 - 540

⇒ 27x3 - 64y3 = 125 + 540

⇒ 27x3 - 64y3 = 665

Hence, the value of 27x3 - 64y3 is 665.

Question 13

If a + b = 8 and ab = 15, find: a3 + b3

Answer

Using the formula,

[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]

So,

⇒ (a + b)3 = a3 + b3 + 3ab(a + b)

Putting the value (a + b) = 8 and ab = 15, we get

⇒ (8)3 = a3 + b3 + 3 ×\times 15 ×\times 8

⇒ 512 = a3 + b3 + 360

⇒ a3 + b3 = 512 - 360

⇒ a3 + b3 = 152

Hence, the value of a3 + b3 is 152.

Question 14

If 3x + 2y = 9 and xy = 3, find: 27x3 + 8y3

Answer

Using the formula,

[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]

So,

⇒ (3x + 2y)3 = (3x)3 + (2y)3 + 3 ×\times 3x ×\times 2y(3x + 2y)

⇒ (3x + 2y)3 = 27x3 + 8y3 + 18xy(3x + 2y)

Putting the value (3x + 2y) = 9 and xy = 3, we get

⇒ (9)3 = 27x3 + 8y3 + 18 ×\times 3 ×\times 9

⇒ 729 = 27x3 + 8y3 + 486

⇒ 27x3 + 8y3 = 729 - 486

⇒ 27x3 + 8y3 = 243

Hence, the value of 27x3 + 8y3 is 243.

Question 15

If 5x - 4y = 7 and xy = 8, find: 125x3 - 64y3

Answer

Using the formula,

[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]

So,

⇒ (5x - 4y)3 = (5x)3 - (4y)3 - 3 ×\times 5x ×\times 4y(5x - 4y)

⇒ (5x - 4y)3 = 125x3 - 64y3 - 60xy(5x - 4y)

Putting the value (5x - 4y) = 7 and xy = 8, we get

⇒ (7)3 = 125x3 - 64y3 - 60 ×\times 8 ×\times 7

⇒ 343 = 125x3 - 64y3 - 3,360

⇒ 125x3 - 64y3 = 343 + 3,360

⇒ 125x3 - 64y3 = 3,703

Hence, the value of 125x3 - 64y3 is 3,703.

Question 16

The difference between two numbers is 5 and their product is 14. Find the difference between their cubes.

Answer

Let the two numbers be x and y.

So,

x - y = 5

And,

xy = 14

Using the formula,

[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]

So,

⇒ (x - y)3 = x3 - y3 - 3xy(x - y)

Putting the value (x - y) = 5 and xy = 14, we get

⇒ (5)3 = x3 - y3 - 3 ×\times 14 ×\times 5

⇒ 125 = x3 - y3 - 210

⇒ x3 - y3 = 125 + 210

⇒ x3 - y3 = 335

Hence, the difference between their cubes is 335.

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