Factorise the following: $a^{3} - b^{3} + 8 c^{3} + 6 a b c$

(a - 2 c) (a^{2} + b^{2} + 4 c^{2} + a b + 2 b c - 2 a c)

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(a - 2 c) (a^{2} + b^{2} + 4 c^{2} + a b + 2 b - 2 a c)

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(a - b + 2 c) (a^{2} + b^{2} + 4 c^{2} + a b + 2 b c - 2 a c)

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(a + b + c) (a^{3} + b^{3} + 8 c^{3} + a b + 2 b - 2 a c)

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Solution

The correct option is D $(a - b + 2 c) (a^{2} + b^{2} + 4 c^{2} + a b + 2 b c - 2 a c)$
$a^{3} - b^{3} + 8 c^{3} + 6 a b c$
Using, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - a c - b c)$
$= (a - b + 2 c) (a^{2} + b^{2} + (2 c)^{2} - a (- b) - a (2 c) - (- b) 2 c)$
$= (a - b + 2 c) (a^{2} + b^{2} + 4 c^{2} + a b - 2 a c + 2 b c)$

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Similar questions

If a, b, c are in G.P., prove that:

(i) $a (b^{2} + c^{2}) = c (a^{2} + b^{2})$

(ii) $a^{2} b^{2} c^{2} (\frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}) = a^{3} + b^{3} + c^{3}$

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