# How do you factor p^6 - 1?

Dec 8, 2015

$\left(p + 1\right) \left(p - 1\right) \left({p}^{2} - p + 1\right) \left({p}^{2} + p + 1\right)$

#### Explanation:

Identities to recognize: $\left\{\begin{matrix}\left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right) \\ \left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right) \\ \left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)\end{matrix}\right.$

Through the first identity, a difference of squares, ${p}^{6} - 1 = \left({p}^{3} + 1\right) \left({p}^{3} - 1\right)$.

Now, we have the other two identities—a sum of cubes and a difference of cubes.

$\textcolor{b l u e}{\left({p}^{3} + 1\right)} \textcolor{m a \ge n t a}{\left({p}^{3} - 1\right)} = \textcolor{b l u e}{\left(p + 1\right) \left({p}^{2} - p + 1\right)} \textcolor{m a \ge n t a}{\left(p - 1\right) \left({p}^{2} + p + 1\right)}$

Thus, ${p}^{6} - 1 = \left(p + 1\right) \left(p - 1\right) \left({p}^{2} - p + 1\right) \left({p}^{2} + p + 1\right)$.