# How do you factor x^(4n) - y^(4n)?

May 12, 2016

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

#### Explanation:

Note that both ${x}^{4 n}$ and ${y}^{4 n}$ are squared terms:

• ${x}^{4 n} = {\left({x}^{2 n}\right)}^{2}$
• ${y}^{4 n} = {\left({y}^{2 n}\right)}^{2}$

With this knowledge, we will factor this expression as a difference of squares:

• ${\textcolor{red}{a}}^{2} - {\textcolor{b l u e}{b}}^{2} = \left(\textcolor{red}{a} + \textcolor{b l u e}{b}\right) \left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right)$

Thus, we see that

${x}^{4 n} - {y}^{4 n}$

$= {\left(\textcolor{red}{{x}^{2 n}}\right)}^{2} - {\left(\textcolor{b l u e}{{y}^{2 n}}\right)}^{2}$

$= \left(\textcolor{red}{{x}^{2 n}} + \textcolor{b l u e}{{y}^{2 n}}\right) \left(\textcolor{red}{{x}^{2 n}} - \textcolor{b l u e}{{y}^{2 n}}\right)$

Notice that we can factor $\left({x}^{2 n} - {y}^{2 n}\right)$ in the same way:

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

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

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