Factorise the following: $27 (x - 1)^{3} + y^{3}$

(3 x - y - 3) (9 x^{2} + 18 x + 9 - 3 x y + 3 y + y^{2})

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(3 x + y - 3) (9 x^{2} - 18 x + 9 - 3 x y + 3 y + y^{2})

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(3 x + y - 3) (x^{2} - 18 x + 9 - 3 x y + 3 y + y^{2})

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(3 x - y - 3) (x^{2} - 18 x + 9 - 3 x y + 3 y + y^{2})

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Solution

The correct option is B $(3 x + y - 3) (9 x^{2} - 18 x + 9 - 3 x y + 3 y + y^{2})$
$27 (x - 1)^{3} + y^{3}$
Let $(x - 1) = a$
Therefore,
$27 a^{3} + y^{3}$
$= (3 a)^{3} + (y)^{3}$
Using, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$= [(3 a + y)] [(3 a)^{2} - (3 a) (y) + (y)^{2}]$
$= (3 a + y) [(9 a)^{2} - 3 a y + (y)^{2}]$
Resubstituting the value we get,
$[3 (x - 1) + y] [9 (x - 1)^{2} - 3 y (x - 1) + y^{2}]$
$= (3 x - 3 + y) [9 (x^{2} - 2 x + 1) - 3 x y + 3 y + y^{2}]$
$= (3 x + y - 3) (9 x^{2} - 18 x + 9 - 3 x y + 3 y + y^{2})$

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