Question

Factorise each of the following:

(i) 8a 3 + b 3 + 12a 2b + 6ab2

(ii) 8a 3 – b 3 – 12a 2b + 6ab2

(iii) 27 – 125a 3 – 135a + 225a 2

(iv) 64a 3 – 27b 3 – 144a 2b + 108ab2

(v) 27p 3 – $\frac{1}{ 216}$ – $\frac{9 }{ 2}$ p2 + $\frac{1 }{4}$ p

Answer 1

It is known that,

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

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

(i) 8a3 + b3 + 12a2b + 6ab2

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

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

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(ii) 8a 3 – b 3 – 12a 2b + 6ab2

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

= (2a - b)3

= (2a - b)(2a - b) (2a - b)

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(iii) 27 – 125a 3 – 135a + 225a 2

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

= (3 - 5a)3 = (3 - 5a) (3 - 5a) (3 - 5a)

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(iv) 64a 3 – 27b 3 – 144a 2b + 108ab2

= (4a)3 - (3b)3 - 3(4a)2(3b) + 3(4a) (3b)2

= (4a - 3b)3 = (4a - 3b) (4a - 3b) (4a - 3b)

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(v) 27p3 – $\frac{1}{ 216}$ – $\frac{9 }{ 2}$ p2 + $\frac{1 }{4}$ p

= (3p)3 - ($\frac{1 }{6}$)3 - 3 (3p)2 ($\frac{1 }{6}$) + 3 (3p) ($\frac{1 }{6}$)2

= (3p - $\frac{1 }{6}$)3

= (3p - $\frac{1 }{6}$)3 (3p - $\frac{1 }{6}$)3 (3p - $\frac{1 }{6}$)3.