# How do you factor x^8 - 1 ?

Oct 8, 2015

$\left({x}^{4} - 1\right) \left({x}^{4} + 1\right)$

#### Explanation:

Using the difference of squares equation, ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
I can determine the answer based on the facts that ${x}^{8}$ can be formatted as a square, ${\left({x}^{4}\right)}^{2}$, and that 1 is a square (for itself). Because of these, I am able to use the difference of squares equation. I take the $a$ value, ${x}^{4}$, and take the $b$ value, 1, and plug them into the equation.

If we wanted to check this, we would need to expand the answer that we have gained using FOIL (First Outer Inner Last).
I will demonstrate the steps below:
$\left({x}^{4} - 1\right) \left({x}^{4} + 1\right)$
First:
${x}^{8} = \left({x}^{4}\right) \left({x}^{4}\right)$
Outer:
${x}^{4} = \left({x}^{4}\right) \left(1\right)$
Inner:
$- {x}^{4} = \left({x}^{4}\right) \left(- 1\right)$
Last:
$- 1 = \left(- 1\right) \left(1\right)$
Final expanded form:
${x}^{8} + {x}^{4} - {x}^{4} - 1$
Which can then cancel the ${x}^{4}$ to get our original equation, confirming the answer is correct.