Factorise: a^3-8 b^3-64 c^3-24 a b c - Vidyadip

Question

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

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Answer

We have,
$a^3-8 b^3-64 c^3-24 a b c$
$=\left\{(a)^3+(-2 b)^3+(-4 c)^3-3(a)(-2 b)(-4 c)\right\}$
$=\{a+(-2 b)+(-4 c)\}\left\{a^2+(-2 b)^2+(-4 c)^2-a(-2 b)-(-2 b)(-4 c)-(-4 c) a\right\}$
${\left[\therefore a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]}$
$=(a+-2 b+-4 c)\left(a^2+4 b^2+16 c^2+2 a b+8 b c+4 c a\right)$

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