# How do you factor (1-x^9)?

Nov 7, 2015

You can use the difference of cubes identity twice to partially factor this as:

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

or more thoroughly to find:

$1 - {x}^{9} = \left(1 - x\right) \left(1 + x + {x}^{2}\right) \left(1 - 2 \cos \left(\frac{2 \pi}{9}\right) x + {x}^{2}\right) \left(1 - 2 \cos \left(\frac{4 \pi}{9}\right) x + {x}^{2}\right) \left(1 - 2 \cos \left(\frac{8 \pi}{9}\right) x + {x}^{2}\right)$

#### Explanation:

The difference of cubes identity is:

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

We can use this identity to partially factor $1 - {x}^{9}$ as follows:

1-x^9 = 1^3-(x^3)^3 = (1-x^3)(1+x^3+(x^3)^2))

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

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

It is possible to factor $1 + {x}^{3} + {x}^{6}$ further into quadratics with Real coefficients, expressible in terms of $\cos$:

$1 + {x}^{3} + {x}^{6} = \left(1 - 2 \cos \left(\frac{2 \pi}{9}\right) x + {x}^{2}\right) \left(1 - 2 \cos \left(\frac{4 \pi}{9}\right) x + {x}^{2}\right) \left(1 - 2 \cos \left(\frac{8 \pi}{9}\right) x + {x}^{2}\right)$

To understand how this works, note that De Moivre's Theorem gives us:

${\left(\cos x + i \sin x\right)}^{n} = \cos \left(n x\right) + i \sin \left(n x\right)$

Putting $n = 9$ and $x = \frac{2 k \pi}{9}$, where $k \in \mathbb{Z}$ we find:

${\left(\cos \left(\frac{2 k \pi}{9}\right) + i \sin \left(\frac{2 k \pi}{9}\right)\right)}^{9} = 1$

This gives us $9$ distinct $9$th roots of $1$.

Every third one of these roots is a cube root of $1$, so a zero of $1 - {x}^{3} = \left(1 - x\right) \left(1 + x + {x}^{2}\right)$. The others are zeros of the remaining factor $\left(1 + {x}^{3} + {x}^{6}\right)$ These $6$ remaining zeros occur in Complex conjugate pairs which combine to give factors with Real coefficients.

For example:

$\left(x - \cos \left(\frac{2 \pi}{9}\right) - i \sin \left(\frac{2 \pi}{9}\right)\right) \left(x - \cos \left(\frac{16 \pi}{9}\right) - i \sin \left(\frac{16 \pi}{9}\right)\right)$

$= \left(x - \cos \left(\frac{2 \pi}{9}\right) - i \sin \left(\frac{2 \pi}{9}\right)\right) \left(x - \cos \left(\frac{2 \pi}{9}\right) + i \sin \left(\frac{2 \pi}{9}\right)\right)$

$= {x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) x + \left({\cos}^{2} \left(\frac{2 \pi}{9}\right) + {\sin}^{2} \left(\frac{2 \pi}{9}\right)\right)$

$= {x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) x + 1$

Similarly for $\frac{4 \pi}{9}$ and $\frac{8 \pi}{9}$