# How do you factor x^3y^6 + 8?

Dec 14, 2015

${x}^{3} {y}^{6} + 8 = \left(x {y}^{2} + 2\right) \left({x}^{2} {y}^{4} - 2 x {y}^{2} + 4\right)$

#### Explanation:

The sum of cubes identity can be written:

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

Both ${x}^{3} {y}^{6} = {\left(x {y}^{2}\right)}^{3}$ and $8 = {2}^{3}$ are perfect cubes, so let $a = x {y}^{2}$ and $b = 2$ to find:

${x}^{3} {y}^{6} + 8$

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

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

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

That's as far as you can get with Real coefficients, but if you allow Complex coefficients this factors a little further:

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

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