# How do you factor x^9 - 27?

Nov 16, 2015

You can use the difference of cubes identity to help factor this, but to find all the factors with Real coefficients requires a little more...

#### Explanation:

You can partially factor ${x}^{9} - 27$ by first recognising that it is a difference of cubes.

The difference of cubes identity can be written:

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

In our case we can put $a = {x}^{3}$ and $b = 3$ to find:

${x}^{9} - 27 = {\left({x}^{3}\right)}^{3} - {3}^{3} = \left({x}^{3} - 3\right) \left({\left({x}^{3}\right)}^{2} + \left({x}^{3}\right) \left(3\right) + {3}^{2}\right)$

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

The remaining cubic factor can itself be treated as a difference of cubes:

${x}^{3} - 3 = {x}^{2} - {\left(\sqrt[3]{3}\right)}^{3} = \left(x - \sqrt[3]{3}\right) \left({x}^{2} + \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}\right)$

The remaining sextic factor can be factored as the product of three quadratics:

${x}^{6} + 3 {x}^{3} + 9 =$

$\left({x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}\right) \cdot$

$\left({x}^{2} - 2 \cos \left(\frac{4 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}\right) \cdot$

$\left({x}^{2} - 2 \cos \left(\frac{8 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}\right)$

To see why requires some Complex arithmetic.

The primitive Complex $9$th root of $1$ is:

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

Then all the $9$th roots of $1$ are $\alpha , {\alpha}^{2} , \ldots , {\alpha}^{9} = 1$

Apart from ${\alpha}^{9} = 1$, these root occur in Complex conjugate pairs, which we can combine to find quadratic factors with Real coefficients. The roots ${\alpha}^{3}$, ${\alpha}^{6}$ and ${\alpha}^{9} = 1$ are all cube roots of $1$ relating to the factors of ${x}^{3} - 3$.

The other $6$ roots combine in pairs to give the three other quadratic factors of the sextic:

$\left(x - \alpha \sqrt[3]{3}\right) \left(x - {\alpha}^{8} \sqrt[3]{3}\right) = {x}^{2} - 2 \cos \left(\frac{2 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}$

$\left(x - {\alpha}^{2} \sqrt[3]{3}\right) \left(x - {\alpha}^{7} \sqrt[3]{3}\right) = {x}^{2} - 2 \cos \left(\frac{4 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}$

$\left(x - {\alpha}^{4} \sqrt[3]{3}\right) \left(x - {\alpha}^{5} \sqrt[3]{3}\right) = {x}^{2} - 2 \cos \left(\frac{8 \pi}{9}\right) \sqrt[3]{3} x + {\sqrt[3]{3}}^{2}$