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Mathematics

Factorise each of the following:

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

(ii) 8a3 - b3 - 12a2b + 6ab2

(iii) 27 - 125a3 - 135a + 225a2

(iv) 64a3 - 27b3 - 144a2b + 108ab2

(v) 27p3 - 1216\dfrac{1}{216} - 92\dfrac{9}{2}p2 + 14\dfrac{1}{4}p

Polynomials

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Answer

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

[∵ a3 + b3 + 3a2b + 3ab2 = (a + b)3]

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

= (2a + b)3

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

(ii) 8a3 - b3 - 12a2b + 6ab2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

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

= (2a - b)3

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

(iii) 27 - 125a3 - 135a + 225a2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

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

= (3 - 5a)3

Hence, 27 - 125a3 - 135a + 225a2 = (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a3 - 27b3 - 144a2b + 108ab2

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

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

= (4a - 3b)3

Hence, 64a3 - 27b3 - 144a2b + 108ab2 = (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p3 - 1216\dfrac{1}{216} - 92\dfrac{9}{2}p2 + 14\dfrac{1}{4}p

[∵ a3 - b3 - 3a2b + 3ab2 = (a - b)3]

= (3p)3 - (16)\Big(\dfrac{1}{6}\Big)3 - 3(3p)2(16)\Big(\dfrac{1}{6}\Big) + 3(3p)(16)\Big(\dfrac{1}{6}\Big)2

= (3p−16)3\Big(3p - \dfrac{1}{6}\Big)^3

Hence, 27p3−1216−92p2+14p=(3p−16)(3p−16)(3p−16)27p^3 - \dfrac{1}{216} - \dfrac{9}{2}p^2 + \dfrac{1}{4}p = \Big(3p - \dfrac{1}{6}\Big)\Big(3p - \dfrac{1}{6}\Big)\Big(3p - \dfrac{1}{6}\Big)

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