# How do you factor completely mn^4 + m^4n?

Dec 4, 2015

Separate out the common factor $m n$ then use the sum of cubes identity to find:

$m {n}^{4} + {m}^{4} n = m n \left(m + n\right) \left({m}^{2} - m n + {n}^{2}\right)$

#### Explanation:

The sum of cubes identity can be written:

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

We can separate out the common factor $m n$ then use the sum of cubes identity with $a = m$ and $b = n$ to find:

$m {n}^{4} + {m}^{4} n = m n \left({n}^{3} + {m}^{3}\right)$

$= m n \left(m + n\right) \left({m}^{2} - m n + {n}^{2}\right)$

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

$= m n \left(m + n\right) \left(m + \omega n\right) \left(m + {\omega}^{2} n\right)$

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