Factorise $27 - x^{3} y^{3} + 6 - 2 x y$

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

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$(2 - x y) (11 - 3 x y + x^{2} y^{2})$

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$(2 - x y) (11 - 4 x y)$

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$(2 + x y) (11 - 3 x y - x^{2} y^{2})$

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Solution

The correct option is B
$(2 - x y) (11 - 3 x y + x^{2} y^{2})$

Given $27 - x^{3} y^{3} + 6 - 2 x y$

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

Using the formula for difference of the cubes , we have
$(3 - x y) (9 - 3 x y + x^{2} y^{2}) + 2 (3 - x y)$
Taking (3-xy ) common
$(2 - x y) (9 - 3 x y + x^{2} y^{2} + 2)$
$(2 - x y) (11 - 3 x y + x^{2} y^{2})$

Suggest Corrections

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