# How do you factor x^9 - x^6 - x^3 + 1?

Mar 11, 2018

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

#### Explanation:

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

Mar 11, 2018

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

#### Explanation:

Let's revise our power rules ( they will come in handy later ):

1. Difference of squares rule: ${x}^{2} - {y}^{2} = \left(x - y\right) \left(x + y\right)$
2. Difference of cubes rule: ${x}^{3} - {y}^{3} = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$
3. Sum of cubes rule: ${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$

${x}^{9} - {x}^{6} - {x}^{3} + 1$

Factorise ${x}^{6}$ from the first two terms,

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

Factorise $\left({x}^{3} - 1\right)$,

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

Apply difference of squares rule to $\left({x}^{6} - 1\right)$,

$\textcolor{red}{\left({x}^{3} + 1\right)} \textcolor{b l u e}{\left({x}^{3} - 1\right)} \textcolor{g r e e n}{\left({x}^{3} - 1\right)}$

Apply sum of cubes rule to $\left({x}^{3} + 1\right)$,

$\textcolor{red}{\left(x + 1\right) \left({x}^{2} - x + 1\right)} \textcolor{b l u e}{\left({x}^{3} - 1\right)} \textcolor{g r e e n}{\left({x}^{3} - 1\right)}$

Apply difference of cubes rule to $\left({x}^{3} - 1\right)$,

$\textcolor{red}{\left(x + 1\right) \left({x}^{2} - x + 1\right)} \textcolor{b l u e}{\left(x - 1\right) \left({x}^{2} + x + 1\right)} \textcolor{g r e e n}{\left(x - 1\right) \left({x}^{2} + x + 1\right)}$

Simplify like polynomials,

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

There you go.

P.S. I did not include the sum of squares rule because that rule dwells into imaginary numbers but if you are keen to know, here it is: ${x}^{2} + {y}^{2} = \left(x + y i\right) \left(x - y i\right)$