Question

factorise a³-8b³-64c³-24

Answer 1

Solution:- (a)³-(2b)³-(4c)³-3(a)(2b)(4c)
[a³-b³-c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=>(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)
=>(a+2b+4c)(a²+4b²+16c²-2ab-8bc-4ca) ANSWER..
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Answer 2

HEY DEAR BRAINLY USER

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UR ANSWER IS =

a³-8b³-64c³-24abc

= a³+(-2b)³+(-4c)³-3*a*-2b*-4c

= (a-2b-4c)(a²+(-2b)²+(-4c)²(-2*a*-2b)(-2*-2b*-4c)(-2*a*-4c))

= (a-2b-4c)(a²+4b²+16c²+4ab-8bc+8ac)

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