Question

factorise:
a³-8b³+64c³+24abc​

Answer 1

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

=[a3-b3-c-3abc=(a+b+c) (a2+b2+c²-ab-bc-ca)

=(a)-(2b)-(4c)3-3(a)(2b)(4c)=(a+2b+4c) [(a)²+(2b)2+(4c)2-(a)(2b)-(2b)(4C)-(4c)(a)

=(a+2b+4c)(a²+4b²+16c²-2ab-8bc-4ca)

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