Factorise:a^{3}-8 b^{3}-64 c^{3}-24 a b c | vedclass

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Question

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$This c

Vedclass · Accepted Answer

$(a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$

Vedclass · Answer

(A) We use the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.Given expression: $a^{3}-8b^{3}-64c^{3}-24abc$This can be rewritten as: $a^{3}+(-2b)^{3}+(-4c)^{3}-3(a)(-2b)(-4c)$.Here,$x=a$,$y=-2b$,and $z=-4c$.Applying the identity:$= (a-2b-4c)((a)^{2}+(-2b)^{2}+(-4c)^{2}-(a)(-2b)-(-2b)(-4c)-(-4c)(a))$$= (a-2b-4c)(a^{2}+4b^{2}+16c^{2}+2ab-8bc+4ca)$.

Factorise:
$a^{3}-8 b^{3}-64 c^{3}-24 a b c$

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Factorise:$a^{3}-8 b^{3}-64 c^{3}-24 a b c$