factorise a² +86² +64 c3 – 24 abc
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Answer:
{(a+2b+4c)(a²+4 b²+16 c²-2 ab-8 bc-4 ca)
Step-by-step explanation:
a³+ 8 b³+64 c³- 24 a bc
using a³+b³ +c³-3 a bc =[ (a+b+c)(a²+ b² + c² -ab -bc - ca)]
identity
(a)³+(2 b)³+(4 c)³-3(a)(2 b)(4 c)
= [(a+2 b+4 c){a² +(2 b)²+(4c)²-a x 2b - 2b x 4c - 4c x a}]
= {(a+2b+4c)(a²+4 b²+16 c²-2 ab-8 bc-4 ca)
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