# How do you factor e^6 + f^3?

Dec 9, 2015

Use the sum of cubes identity to find:

${e}^{6} + {f}^{3} = \left({e}^{2} + f\right) \left({e}^{4} - {e}^{2} f + {f}^{2}\right)$

#### Explanation:

The sum of cubes identity may be written:

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

If we let $a = {e}^{2}$ and $b = f$ then we find:

${e}^{6} + {f}^{3} = {\left({e}^{2}\right)}^{3} + {f}^{3}$

$= \left({e}^{2} + f\right) \left({\left({e}^{2}\right)}^{2} - \left({e}^{2}\right) f + {f}^{2}\right)$

$= \left({e}^{2} + f\right) \left({e}^{4} - {e}^{2} f + {f}^{2}\right)$

If we allow Complex coefficients then this can be factored further:

$= \left({e}^{2} + f\right) \left({e}^{2} + \omega f\right) \left({e}^{2} + {\omega}^{2} f\right)$

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