Factorise : $8 x^{3} - 27 y^{3} - 2 x + 3 y :$

(2 x + 3 y) (4 x^{2} - x y + 9 y^{2} + 1)

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(2 x + y) (4 x^{2} - 6 x y - 9 y^{2})

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(x - 3 y) (4 x^{2} - 3 x y + 9 y^{2})

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(2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2} - 1)

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Solution

The correct option is D $(2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2} - 1)$
Given, $8 x^{3} - 27 y^{3} - 2 x + 3 y$
Rearranging the terms , we get
$(2 x)^{3} - (3 y)^{3} - (2 x - 3 y)$
$= (2 x - 3 y) [(2 x)^{2} + (2 x) (3 y) + (3 y)^{2}] - (2 x - 3 y)$
$= (2 x - 3 y) [4 x^{2} + 6 x y + 9 y^{2}] - (2 x - 3 y)$
$= (2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2} - 1)$

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x = 1, y = 1

(2 x + 3 y) (4 x^{2} - 6 x y + 9 y^{2})

(2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2})

(2 x - 3 y) (2 x + 3 y) (4 x^{2} + 9 y^{2})

(2 x - 3 y) (2 x + 3 y) ({4 x}^{2} + {9 y}^{2})

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

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