# How do you factor ((x-y)^3) +8?

Aug 18, 2016

${\left(x - y\right)}^{3} + 8$

$= \left(x - y + 2\right) \left({x}^{2} - 2 x y + {y}^{2} - 2 x + 2 y + 4\right)$

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

#### Explanation:

The sum of cubes identity can be written:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

Use this with $a = \left(x - y\right)$ and $b = 2$ to find:

${\left(x - y\right)}^{3} + 8$

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

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

$= \left(x - y + 2\right) \left({x}^{2} - 2 x y + {y}^{2} - 2 x + 2 y + 4\right)$

If you allow Complex coefficients, then this factors further as:

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

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$

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