How do you factor $x^{3} y^{3} + z^{3}$ $x^3y^3 + z^3$ ?

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Answer 1 · 2015-07-22T07:39:48

Use sum of cubes identity to find:

$x^{3} y^{3} + z^{3} = (x y + z) (x^{2} y^{2} - x y z + z^{2})$ $x^3y^3+z^3 = (xy+z)(x^2y^2-xyz+z^2)$

Use the sum of cubes identity:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$

with $a = x y$ $a=xy$ and $b = z$ $b=z$ as follows:

$x^{3} y^{3} + z^{3}$ $x^3y^3+z^3$

$= {(x y)}^{3} + z^{3}$ $=(xy)^3+z^3$

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

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

How do you factor x3y3+z3x^3y^3 + z^3?