# How do you factor u^3 - 1?

Sep 23, 2015

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

#### Explanation:

There is a useful 'difference of cubes' identity:

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

In our example, $a = u$ and $b = 1$ :

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

In fact, in general we find:

${a}^{n} - {b}^{n} = \left(a - b\right) \left({a}^{n - 1} + {a}^{n - 2} b + {a}^{n - 3} {b}^{2} + \ldots + {b}^{n - 1}\right)$

When you multiply it out, most of the terms cancel.

When $n$ is even, the second factor can be factorised further:

${a}^{n - 1} + {a}^{n - 2} b + {a}^{n - 3} {b}^{2} + \ldots + {b}^{n - 1}$

$= \left(a + b\right) \left({a}^{n - 2} + {a}^{n - 4} {b}^{2} + \ldots + {b}^{n - 2}\right)$

There are no more linear factors with Real coefficients, but there are quadratic factors with Real coefficients.

For example:

${a}^{6} - {b}^{6} = \left(a - b\right) \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right) \left({a}^{2} + a b + {b}^{2}\right)$