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
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
Solution and Explanation
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)
(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)
(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)
(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)
(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.
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