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Chapter 5

Factorisation — Exercise 5(D)

Class - 9 Concise Mathematics Selina

Exercise 5(D)

Question 1(a)

a3 - 8 in the form of factors is :

  1. (a - 2)(a2 - 2a + 4)

  2. (a - 2)(a2 + 2a + 4)

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

  4. (a + 2)(a2 - 2a + 4)

Answer

Given,

   a3 - 8

= (a)3 - (2)3

= (a - 2)[(a)2 + 2 × a + (2)2]

= (a - 2)(a2 + 2a + 4)

Hence, Option 2 is the correct option.

Question 1(b)

27 + 8x3 in the form of factors is :

  1. (3 + 2x)(9 + 6x + 4x2)

  2. (3 - 2x)(9 + 6x + 4x2)

  3. (3 + 2x)(9 - 6x + 4x2)

  4. (3 - 2x)(9 - 6x + 4x2)

Answer

Given,

   27 + 8x3

= (3)3 + (2x)3

= (3 + 2x)[(3)2 - 3 × 2x + (2x)2]

= (3 + 2x)(9 - 6x + 4x2)

Hence, Option 3 is the correct option.

Question 1(c)

8a3 + 1 in the form of factors is :

  1. (2a + 1)(4a2 - 2a + 1)

  2. (2a - 1)(4a2 - 2a + 1)

  3. (2a + 1)(4a2 + 2a + 1)

  4. (2a - 1)(4a2 + 2a + 1)

Answer

Given,

   8a3 + 1

= (2a)3 + (1)3

= (2a + 1)[(2a)2 - 2a × 1 + (1)2]

= (2a + 1)(4a2 - 2a + 1)

Hence, Option 1 is the correct option.

Question 1(d)

(x38−y327)\Big(\dfrac{x^3}{8} - \dfrac{y^3}{27}\Big) in the form of factors is :

  1. (x2−y3)(x24−xy6+y29)\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)

  2. (x2+y3)(x24−xy6+y29)\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)

  3. (x2−y3)(x24+xy6+y29)\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)

  4. (x2+y3)(x24+xy6+y29)\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)

Answer

Given,

=(x38−y327)=(x2)3−(y3)3=(x2−y3)[(x2)2+x2×y3+(y3)2]=(x2−y3)(x24+xy6+y29).\phantom{=} \Big(\dfrac{x^3}{8} - \dfrac{y^3}{27}\Big) \\[1em] = \Big(\dfrac{x}{2}\Big)^3 - \Big(\dfrac{y}{3}\Big)^3 \\[1em] = \Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big[\Big(\dfrac{x}{2}\Big)^2 + \dfrac{x}{2} \times \dfrac{y}{3} + \Big(\dfrac{y}{3}\Big)^2\Big] \\[1em] = \Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big).

Hence, Option 3 is the correct option.

Factorise :

Question 2

64 - a3b3

Answer

Given,

   64 - a3b3

= (4)3 - (ab)3

= (4 - ab)[(4)2 + 4 × ab + (ab)2]

= (4 - ab)(16 + 4ab + a2b2).

Hence, 64 - a3b3 = (4 - ab)(16 + 4ab + a2b2).

Question 3

a6 + 27b3

Answer

Given,

   a6 + 27b3

= (a2)3 + (3b)3

= (a2 + 3b)[(a2)2 - a2 × 3b + (3b)2]

= (a2 + 3b)(a4 - 3a2b + 9b2).

Hence, a6 + 27b3 = (a2 + 3b)(a4 - 3a2b + 9b2).

Question 4

3x7y - 81x4y4

Answer

Given,

   3x7y - 81x4y4

= 3x4y[x3 - 27y3]

= 3x4y[(x)3 - (3y)3]

= 3x4y(x - 3y)[(x)2 + x × 3y + (3y)2]

= 3x4y(x - 3y)(x2 + 3xy + 9y2).

Hence, 3x7y - 81x4y4 = 3x4y(x - 3y)(x2 + 3xy + 9y2).

Question 5

a3 - 27a3\dfrac{27}{a^3}

Answer

Given,

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

Hence, a3−27a3=(a−3a)(a2+3+9a2).a^3 - \dfrac{27}{a^3} = \Big(a - \dfrac{3}{a}\Big)\Big(a^2 + 3 + \dfrac{9}{a^2}\Big).

Question 6

a3 + 0.064

Answer

Given,

    a3 + 0.064

= (a)3 + (0.4)3

= (a + 0.4)[a2 - 0.4 × a + (0.4)2]

= (a + 0.4)(a2 - 0.4a + 0.16)

Hence, a3 + 0.064 = (a + 0.4)(a2 - 0.4a + 0.16)

Question 7

(x - y)3 - 8x3

Answer

Given,

    (x - y)3 - 8x3

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

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

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

= -(x + y)(7x2 - 4xy + y2).

Hence, (x - y)3 - 8x3 = -(x + y)(7x2 - 4xy + y2).

Question 8

(8a327−b38)\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big)

Answer

Given,

=(8a327−b38)=(2a3)3−(b2)3=(2a3−b2)[(2a3)2+2a3×b2+(b2)2]=(2a3−b2)(4a29+ab3+b24)\phantom{=}\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) \\[1em] = \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \dfrac{2a}{3} \times \dfrac{b}{2} + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big) \\[1em]

Hence, (8a327−b38)=(2a3−b2)(4a29+ab3+b24).\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).

Question 9

a6 - b6

Answer

Given,

    a6 - b6

= (a3)2 - (b3)2

= (a3 + b3)(a3 - b3)

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

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

Hence, a6 - b6 = (a + b)(a - b)(a2 - ab + b2)(a2 + ab + b2).

Question 10

a6 - 7a3 - 8

Answer

Given,

    a6 - 7a3 - 8

= (a3)2 - 7a3 - 8

Substituting a3 = x, we get :

= x2 - 7x - 8

= x2 - 8x + x - 8

= x(x - 8) + 1(x - 8)

= (x - 8)(x + 1)

= (a3 - 8)(a3 + 1)

= (a3 - 23)(a3 + 13)

= (a - 2)(a2 + a × 2 + 22)(a + 1)(a2 - a × 1 + 12)

= (a - 2)(a2 + 2a + 4)(a + 1)(a2 - a + 1)

= (a - 2)(a + 1)(a2 + 2a + 4)(a2 - a + 1).

Hence, a6 - 7a3 - 8 = (a - 2)(a + 1)(a2 + 2a + 4)(a2 - a + 1).

Question 11

a3 - 27b3 + 2a2b - 6ab2

Answer

Given,

    a3 - 27b3 + 2a2b - 6ab2

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

= (a - 3b)[(a)2 + a × 3b + (3b)2] + 2ab(a - 3b)

= (a - 3b)[a2 + 3ab + 9b2] + 2ab(a - 3b)

= (a - 3b)(a2 + 3ab + 9b2 + 2ab)

= (a - 3b)(a2 + 5ab + 9b2).

Hence, a3 - 27b3 + 2a2b - 6ab2 = (a - 3b)(a2 + 5ab + 9b2).

Question 12

8a3 - b3 - 4ax + 2bx

Answer

Given,

    8a3 - b3 - 4ax + 2bx

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

= (2a - b)[(2a)2 + 2a × b + (b)2] - 2x(2a - b)

= (2a - b)(4a2 + 2ab + b2) - 2x(2a - b)

= (2a - b)(4a2 + 2ab + b2 - 2x).

Hence, 8a3 - b3 - 4ax + 2bx = (2a - b)(4a2 + 2ab + b2 - 2x).

Question 13

a - b - a3 + b3

Answer

Given,

    a - b - a3 + b3

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

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

= (a - b)(1 - a2 - ab - b2).

Hence, a - b - a3 + b3 = (a - b)(1 - a2 - ab - b2).

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