Factorise: $x^{3} + 6 x^{2} y + 12 x y^{2} + 9 y^{3}$

(x + 3 y) (x^{2} + y^{2} + 3 x^{2} y)

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(x + 3 y) (x^{2} + 3 y^{2} + 3 x y)

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(x - 3 y) (x^{2} + y^{2} + 3 x)

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(x - 3 y) (x^{2} + 3 y^{2} + 3 y)

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Solution

$x^{3} + 6 x^{2} y + 12 x y^{2} + 9 y^{3}$
$= x^{3} + 6 x^{2} y + 12 x y^{2} + 8 y^{3} + y^{3}$
$= {x^{3} + 3 x^{2} (2 y) + 3 (x) (2 y)^{2} + (2 y)^{3}} + y^{3}$
$= (x + 2 y)^{3} + y^{3}$ [Since, $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = (a + b)^{3}$ ]
$= (x + 2 y + y) ((x + 2 y)^{2} - (x + 2 y) y + y^{2})$ [Since $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ ]
$= (x + 3 y) (x^{2} + 2 x (2 y) + (2 y)^{2} - x y - 2 y^{2} + y^{2})$
$= (x + 3 y) (x^{2} + 4 x y + 4 y^{2} - x y - 2 y^{2} + y^{2})$
$= (x + 3 y) (x^{2} + 3 y^{2} + 3 x y)$
Hence, the correct option is B.

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Question 1 (v)

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