# How do you factor completely x^9-1?

Apr 6, 2016

${x}^{9} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

#### Explanation:

To factorize ${x}^{9} - 1$, we can use the identity

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

Hence ${x}^{9} - 1 = {\left({x}^{3}\right)}^{3} - {1}^{3} = \left({x}^{3} - 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

factorizing ${x}^{3} - 1$ further

${x}^{9} - 1 = \left({x}^{3} - 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

= $\left(x - 1\right) \left({x}^{2} + x + 1\right) \left({x}^{6} + {x}^{3} + 1\right)$

Apr 6, 2016

The separate factors are (x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8. The first factor $\left(x - 1\right)$ is the only real factor.

#### Explanation:

The roots of this equation ${x}^{9} - 1 = 0$ are the 9 values of $x = {1}^{\frac{1}{9}}$.
They are$\left(\cos \left(2 k \frac{\pi}{9}\right) + i \sin \left(2 k \frac{\pi}{9}\right)\right) , k = 0 , 1 , 2 , 3 , \ldots , 8$.
k=0 gives the only real root 1.

So, the factors of ${x}^{0} - 1$ are (x-(cos(2kpi/9)+i sin (2kpi/9)), k=0, 1, 2, 3, ..., 8.