How do you factor $x^{3} y^{3} - 8$ $x^3y^3 - 8$ ?

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How do you factor $x^{3} y^{3} - 8$ $x^3y^3 - 8$ ?

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Answer 1 · 2016-08-04T12:08:27

$= (x y - 2) (x^{2} y^{2} + 2 x y + 2)$ $=(xy-2)(x^2y^2+2xy+2)$

Explanation:

$x^{3} y^{3} - 8$ $x^3y^3-8$
or
${(x y)}^{3} - 2^{3}$ $(xy)^3-2^3$
Since $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$
or
$= (x y - 2) ({(x y)}^{2} + 2 x y + 2)$ $=(xy-2)((xy)^2+2xy+2)$
$= (x y - 2) (x^{2} y^{2} + 2 x y + 2)$ $=(xy-2)(x^2y^2+2xy+2)$

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How do you factor $x^{3} y^{3} - 8$ $x^3y^3 - 8$ ?

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