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Question
Factorise:
a3 – 8b3 – 64c3 – 24abc
Solution
a3 – 8b3 – 64c3 – 24abc = (a)3 + (–2b)3 + (–4c)3 – 3 × (a) × (–2b) × (–4c)
= (a – 2b – 4c)[(a)2 + (–2b)2 + (–4c)2 – a(–2b) – (–2b)(–4c) – (–4c)(a)] ...[Using identity, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)]
= (a – 2b – 4c)(a2 + 4b2 + 16c2 + 2ab – 8bc + 4ac)
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