Factorise: $8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}$

(2a+b)(2a+b)(2a+b)(2a+b)(2a+b)(2a+b)

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(2ab)(2a+b)(2a+b)(2ab)(2a+b)(2a+b)

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(2a-b)(2a+b)

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(2a+b)(2a+b)

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Solution

The correct option is A
(2a+b)(2a+b)(2a+b)

Since $x^{3} + y^{3} + 3 x^{2} y + 3 x y^{2} = (x + y)^{3}$
$8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}$

$= (2 a)^{3} + (b)^{3} + 3 (2 a)^{2} (b) + 3 (2 a) (b)^{2}$

$= (2 a + b)^{3}$

$= (2 a + b) (2 a + b) (2 a + b)$

Hence factorised.

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Similar questions

Factorise : 8a³ + b³ + 12a²b + 6ab²

8 a^{3} + b^{3} + 12 a^{2} b + 6 a b^{2}

= (2 a + b) (2 a + b) (2 a + b)

8 a^{3} - b^{3} - 12 a^{2} b + 6 a b^{2}

= (2 a - b) (2 a - b) (2 a - b)

Factorise:
$a^{8} - 2 a^{4} b^{4} + b^{8}$

y = a x^{2} + b x + c

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