# How do you factor completely x^3 - 1?

Using the identity ${a}^{3} - {b}^{3} = \left(a - b\right) \cdot \left({a}^{2} + a b + {b}^{2}\right)$ we have that

${x}^{3} - 1 = \left(x - 1\right) \cdot \left({x}^{2} + x + 1\right)$

Jan 3, 2016

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

#### Explanation:

This is a difference of cubes.

Differences of cubes can be factored as follows:

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

In your situation, $a = x$ and $b = 1$, since ${a}^{3} = {x}^{3}$ and ${b}^{3} = 1$.

${x}^{3} - {1}^{3} = \left(x - 1\right) \left({x}^{2} + x \left(1\right) + {1}^{2}\right)$

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