# How do you factor x^3 (2x - y) + xy(2x - y)^2?

May 4, 2016

${x}^{3} \left(2 x - y\right) + x y {\left(2 x - y\right)}^{2} = \left(2 x - y\right) x \left(x - \left(1 + \sqrt{2}\right) y\right) \left(x - \left(1 - \sqrt{2}\right) y\right)$

#### Explanation:

Use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - y\right)$ and $b = \sqrt{2} y$ as follows:

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

$= \left(2 x - y\right) \left({x}^{3} + x y \left(2 x - y\right)\right)$

$= \left(2 x - y\right) \left({x}^{3} - 2 {x}^{2} y - x {y}^{2}\right)$

$= \left(2 x - y\right) x \left({x}^{2} - 2 x y - {y}^{2}\right)$

$= \left(2 x - y\right) x \left({\left(x - y\right)}^{2} - 2 {y}^{2}\right)$

$= \left(2 x - y\right) x \left({\left(x - y\right)}^{2} - {\left(\sqrt{2} y\right)}^{2}\right)$

$= \left(2 x - y\right) x \left(\left(x - y\right) - \sqrt{2} y\right) \left(\left(x - y\right) + \sqrt{2} y\right)$

$= \left(2 x - y\right) x \left(x - \left(1 + \sqrt{2}\right) y\right) \left(x - \left(1 - \sqrt{2}\right) y\right)$