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Chapter 4

Factorisation — Exercise 4.5

Class - 9 ML Aggarwal Understanding ICSE Mathematics

Exercise 4.5

Question 1(i)

Factorise the following:

8x3 + y3

Answer

8x3 + y3 = (2x)3 + (y)3

We know that,

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

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

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

Hence, 8x3 + y3 = (2x + y)(4x2 - 2xy + y2).

Question 1(ii)

Factorise the following:

64x3 - 125y3

Answer

64x3 - 125y3 = (4x)3 - (5y)3.

We know that,

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

∴ (4x)3 - (5y)3 = (4x - 5y)[(4x)2 + 4x(5y) + (5y)2]

= (4x - 5y)(16x2 + 20xy + 25y2).

Hence, 64x3 - 125y3 = (4x - 5y)(16x2 + 20xy + 25y2).

Question 2(i)

Factorise the following:

64x3 + 1

Answer

64x3 + 1 = (4x)3 + (1)3.

We know that,

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

∴ (4x)3 + (1)3 = (4x + 1)[(4x)2 - 4x.1 + 12]

= (4x + 1)(16x2 - 4x + 1).

Hence, 64x3 + 1 = (4x + 1)(16x2 - 4x + 1).

Question 2(ii)

Factorise the following:

7a3 + 56b3

Answer

7a3 + 56b3 = 7(a3 + 8b3)

= 7[(a)3 + (2b)3]

We know that,

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

∴ 7[(a)3 + (2b)3] = 7(a + 2b)[a2 - a.2b + (2b)2]

= 7(a + 2b)(a2 - 2ab + 4b2).

Hence, 7a3 + 56b3 = 7(a + 2b)(a2 - 2ab + 4b2).

Question 3(i)

Factorise the following:

x6343+343x6\dfrac{x^6}{343} + \dfrac{343}{x^6}.

Answer

x6343+343x6=(x27)3+(7x2)3\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3.

We know that,

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

∴(x27)3+(7x2)3=(x27+7x2)[(x27)2−x27×7x2+(7x2)2]=(x27+7x2)(x449−1+49x4).\therefore \Big(\dfrac{x^2}{7}\Big)^3 + \Big(\dfrac{7}{x^2}\Big)^3 = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big[\Big(\dfrac{x^2}{7}\Big)^2 - \dfrac{x^2}{7} \times \dfrac{7}{x^2} + \Big(\dfrac{7}{x^2}\Big)^2\Big] \\[1em] = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).

Hence, x6343+343x6=(x27+7x2)(x449−1+49x4).\dfrac{x^6}{343} + \dfrac{343}{x^6} = \Big(\dfrac{x^2}{7} + \dfrac{7}{x^2}\Big)\Big(\dfrac{x^4}{49} - 1 + \dfrac{49}{x^4}\Big).

Question 3(ii)

Factorise the following:

8x3−127y38x^3 - \dfrac{1}{27y^3}.

Answer

8x3−127y3=(2x)3−(13y)38x^3 - \dfrac{1}{27y^3} = (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3

We know that,

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

∴(2x)3−(13y)3=(2x−13y)[(2x)2+2x.13y+(13y)2]=(2x−13y)(4x2+2x3y+19y2).\therefore (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3 = \Big(2x - \dfrac{1}{3y}\Big)\Big[(2x)^2 + 2x.\dfrac{1}{3y} + \Big(\dfrac{1}{3y}\Big)^2\Big] \\[1em] = \Big(2x - \dfrac{1}{3y}\Big)\Big(4x^2 + \dfrac{2x}{3y} + \dfrac{1}{9y^2}\Big).

Hence, 8x3−127y3=(2x−13y)(4x2+2x3y+19y2).8x^3 - \dfrac{1}{27y^3} = \Big(2x - \dfrac{1}{3y}\Big)\Big(4x^2 + \dfrac{2x}{3y} + \dfrac{1}{9y^2}\Big).

Question 4(i)

Factorise the following:

x2 + x5

Answer

x2 + x5 = x2(1 + x3) = x2[(1)3 + (x)3].

We know that,

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

x2[(1)3 + (x)3] = x2(1 + x)(12 - 1.x + x2)

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

Hence, x2 + x5 = x2(1 + x)(1 - x + x2).

Question 4(ii)

Factorise the following:

32x4 - 500x

Answer

32x4 - 500x = 4x(8x3 - 125) = 4x[(2x)3 - 53]

We know that,

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

4x[(2x)3 - 53] = 4x(2x - 5)[(2x)2 + 2x.5 + (5)2]

= 4x(2x - 5)(4x2 + 10x + 25).

Hence, 32x4 - 500x = 4x(2x - 5)(4x2 + 10x + 25).

Question 5(i)

Factorise the following:

27x3y3 - 8

Answer

27x3y3 - 8 = (3xy)3 - (2)3.

We know that,

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

(3xy)3 - (2)3 = (3xy - 2)[(3xy)2 + 3xy.2 + 22]

= (3xy - 2)(9x2y2 + 6xy + 4).

Hence, 27x3y3 - 8 = (3xy - 2)(9x2y2 + 6xy + 4).

Question 5(ii)

Factorise the following:

27(x + y)3 + 8(2x - y)3

Answer

27(x + y)3 + 8(2x - y)3 = [3(x + y)]3 + [2(2x - y)]3

We know that,

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

∴ [3(x + y)]3 + [2(2x - y)]3 = [3(x + y) + 2(2x - y)][{3(x + y)}2 - 3(x + y) × 2(2x - y) + {2(2x - y)}2]

= [3x + 3y + 4x - 2y][9(x2 + y2 + 2xy) - 6(x + y)(2x - y) + 4(4x2 + y2 - 4xy)]

= (7x + y)[9x2 + 9y2 + 18xy - 6(2x2 - xy + 2xy - y2) + 16x2 + 4y2 - 16xy]

= (7x + y)[9x2 + 9y2 + 18xy - 12x2 + 6xy - 12xy + 6y2 + 16x2 + 4y2 - 16xy]

= (7x + y)(13x2 + 19y2 - 4xy).

Hence, 27(x + y)3 + 8(2x - y)3 = (7x + y)(13x2 + 19y2 - 4xy).

Question 6(i)

Factorise the following:

a3 + b3 + a + b

Answer

We know that,

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

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

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

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

Question 6(ii)

Factorise the following:

a3 - b3 - a + b

Answer

We know that,

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

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

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

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

Question 7(i)

Factorise the following:

x3 + x + 2

Answer

x3 + x + 2 = x3 + x + 1 + 1.

= x3 + 1 + x + 1

= x3 + 13 + x + 1

We know that,

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

x3 + 13 + x + 1 = (x + 1)(x2 - x + 1) + (x + 1)

= (x + 1)(x2 - x + 1 + 1)

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

Hence, x3 + x + 2 = (x + 1)(x2 - x + 2).

Question 7(ii)

Factorise the following:

a3 - a - 120

Answer

a3 - a - 120 = a3 - a - 125 + 5.

Rearranging above terms we get,

a3 - 125 - a + 5

= [a3 - (5)3] - (a - 5)

We know that,

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

[a3 - (5)3] - (a - 5) = (a - 5)(a2 + 5a + 52) - (a - 5)

= (a - 5)(a2 + 5a + 25 - 1)

= (a - 5)(a2 + 5a + 24).

Hence, a3 - a - 120 = (a - 5)(a2 + 5a + 24).

Question 7(iii)

a3 - 2a - 115

Answer

Given,

⇒ a3 - 2a - 115

⇒ a3 - 2a - 125 + 10

Rearranging above terms we get,

⇒ a3 - 125 - 2a + 10

⇒ (a3 - 53) - 2(a - 5)

We know that,

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

⇒ (a - 5)(a2 + 5a + 52) - 2(a - 5)

⇒ (a - 5)[(a2 + 5a + 52) - 2]

⇒ (a - 5)[a2 + 5a + 25 - 2]

⇒ (a - 5)(a2 + 5a + 23).

Hence, a3 - 2a - 115 = (a - 5)(a2 + 5a + 23).

Question 8(i)

Factorise the following:

x3 + 6x2 + 12x + 16

Answer

x3 + 6x2 + 12x + 16 can be written as x3 + 6x2 + 12x + 8 + 8.

Above terms can be written as,

[(x)3 + (3 × 2 × x2) + (3 × 22 × x) + 23] + 8

= [(x)3 + 3 × x × 2(x + 2) + 23] + 23 ........(i)

We know that,

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

Comparing equation (i) and (ii) we get,

a = x and b = 2.

∴ [(x)3 + 3 × x × 2(x + 2) + 23] + 23 = (x + 2)3 + 23.

We know that,

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

∴ (x + 2)3 + 23 = (x + 2 + 2)[(x + 2)2 - (x + 2).2 + 22]

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

= (x + 4)(x2 + 2x + 4)

Hence, x3 + 6x2 + 12x + 16 = (x + 4)(x2 + 2x + 4).

Question 8(ii)

Factorise the following:

a3 - 3a2b + 3ab2 - 2b3

Answer

a3 - 3a2b + 3ab2 - 2b3 = a3 - 3a2b + 3ab2 - b3 - b3.

We know that,

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

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

We know that,

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

∴ (a - b)3 - b3 = (a - b - b)[(a - b)2 + (a - b).b + b2]

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

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

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

Question 9(i)

Factorise the following:

2a3 + 16b3 - 5a - 10b

Answer

2a3 + 16b3 - 5a - 10b = 2(a3 + 8b3) - 5(a + 2b)

= 2[a3 + (2b)3] - 5(a + 2b)

We know that,

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

∴ 2[a3 + (2b)3] - 5(a + 2b) = 2(a + 2b)(a2 - 2ab + 4b2) - 5(a + 2b)

= (a + 2b)[2(a2 - 2ab + 4b2) - 5]

= (a + 2b)(2a2 - 4ab + 8b2 - 5).

Hence, 2a3 + 16b3 - 5a - 10b = (a + 2b)(2a2 - 4ab + 8b2 - 5).

Question 9(ii)

Factorise the following:

a3−1a3−2a+2aa^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a}.

Answer

a3−1a3−2a+2a=a3−1a3−2(a−1a)a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} = a^3 - \dfrac{1}{a^3} - 2\Big(a - \dfrac{1}{a}\Big).

We know that,

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

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

Hence, a3−1a3−2a+2a=(a−1a)(a2+1a2−1).a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} = \Big(a - \dfrac{1}{a}\Big)\Big(a^2 + \dfrac{1}{a^2} - 1\Big).

Question 10(i)

Factorise the following:

a6 - b6

Answer

a6 - b6 = (a2)3 - (b2)3.

We know that,

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

∴ (a2)3 - (b2)3 = (a2 - b2)[(a2)2 + a2b2 + (b2)2]

= (a2 - b2)(a4 + a2b2 + b4)

= (a - b)(a + b)(a4 + a2b2 + b4)

= (a - b)(a + b)[(a2)2 + 2a2b2 + (b2)2 - a2b2]

=(a - b)(a + b)[(a2 + b2)2 - a2b2]

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

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

Question 10(ii)

Factorise the following:

x6 - 1

Answer

x6 - 1 = (x3)2 - (1)2.

We know that,

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

∴ (x3)2 - (1)2 = (x3 - 1)(x3 + 1).

We know that,

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

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

∴ (x3 - 1)(x3 + 1) = (x - 1)(x2 + x + 1)(x + 1)(x2 - x + 1)

Hence, x6 - 1 = (x - 1)(x + 1)(x2 + x + 1)(x2 - x + 1).

Question 11(i)

Factorise the following:

64x6 - 729y6

Answer

64x6 - 729y6 = [(2x)3]2 - [(3y)3]2.

We know that,

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

∴ [(2x)3]2 - [(3y)3]2 = [{(2x)3 - (3y)3}{(2x)3 + (3y)3}]

We know that,

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

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

∴ [{(2x)3 - (3y)3}{(2x)3 + (3y)3}] = (2x - 3y)[(2x)2 + 2x.3y + (3y)2](2x + 3y)[(2x)2 - 2x.3y + (3y)2]

= (2x - 3y)(4x2 + 6xy + 9y2)(2x + 3y)(4x2 - 6xy + 9y2)

Hence, 64x6 - 729y6 = (2x - 3y)(2x + 3y)(4x2 + 6xy + 9y2)(4x2 - 6xy + 9y2).

Question 11(ii)

Factorise the following:

x2−8xx^2 - \dfrac{8}{x}

Answer

Above terms can be written as,

⇒x2−8x=1x(x3−8)=1x(x3−23).\Rightarrow x^2 - \dfrac{8}{x} = \dfrac{1}{x}\Big(x^3 - 8\Big) \\[1em] = \dfrac{1}{x}(x^3 - 2^3).

We know that,

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

=1x(x3−23)= \dfrac{1}{x}(x^3 - 2^3)

We know that,

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

1x(x3−23)=1x(x−2)(x2+2×x+22)=1x(x−2)(x2+2x+4).\dfrac{1}{x}(x^3 - 2^3) = \dfrac{1}{x}(x - 2)(x^2 + 2 \times x + 2^2) \\[1em] = \dfrac{1}{x}(x - 2)(x^2 + 2x + 4).

Hence, x2−8x=1x(x−2)(x2+2x+4).x^2 - \dfrac{8}{x} = \dfrac{1}{x}(x - 2)(x^2 + 2x + 4).

Question 12(i)

Factorise the following:

250(a - b)3 + 2

Answer

250(a - b)3 + 2 = 2[125(a - b)3 + 1]

= 2[{5(a - b)}3 + 13]

We know that,

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

∴ 2[{5(a - b)}3 + 13] = 2[(5a - 5b + 1){(5a - 5b)2 - (5a - 5b)1 + 12}]

= 2(5a - 5b + 1)(25a2 + 25b2 - 50ab - 5a + 5b + 1).

Hence, 250(a - b)3 + 2 = 2(5a - 5b + 1)(25a2 + 25b2 - 50ab - 5a + 5b + 1).

Question 12(ii)

Factorise the following:

32a2x3 - 8b2x3 - 4a2y3 + b2y3.

Answer

32a2x3 - 8b2x3 - 4a2y3 + b2y3 = 8x3(4a2 - b2) - y3(4a2 - b2)

= (4a2 - b2)(8x3 - y3)

= [(2a)2 - (b)2][(2x)3 - (y)3]

We know that,

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

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

∴ [(2a)2 - (b)2][(2x)3 - (y)3] = (2a - b)(2a + b)(2x - y)[(2x)2 + 2xy + y2]

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

Hence, 32a2x3 - 8b2x3 - 4a2y3 + b2y3 = (2a - b)(2a + b)(2x - y)(4x2 + 2xy + y2).

Question 13(i)

Factorise the following:

x9 + y9

Answer

x9 + y9 = (x3)3 + (y3)3

We know that,

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

(x3)3 + (y3)3 = (x3 + y3)[(x3)2 - x3y3 + (y3)2]

= (x3 + y3)(x6 - x3y3 + y6)

= (x + y)(x2 - xy + y2)(x6 - x3y3 + y6)

Hence, x9 + y9 = (x + y)(x2 - xy + y2)(x6 - x3y3 + y6).

Question 13(ii)

Factorise the following:

x6 - 7x3 - 8

Answer

Factorising x6 - 7x3 - 8 we get,

x6 - 7x3 - 8 = x6 - 8x3 + x3 - 8

= x3(x3 - 8) + 1(x3 - 8)

= (x3 - 8)(x3 + 1)

= [(x)3 - (2)3][(x)3 + 13]

We know that,

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

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

∴ x3 + 13 = (x + 1)(x2 - x + 1)

and,

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

∴ [(x)3 - (2)3][(x)3 + 13] = (x - 2)(x2 + 2x + 4)(x + 1)(x2 - x + 1).

Hence, x6 - 7x3 - 8 = (x - 2)(x2 + 2x + 4)(x + 1)(x2 - x + 1).

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