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

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

#### Explanation:

Start from the given:

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

by grouping method

first two terms, factor ${x}^{6}$ and last two terms, factor the $- 1$

that is

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

factor out the common binomial factor $\left({x}^{3} - 1\right)$ so that

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

at this point, use " sum or difference of two cubes" forms
and difference of two squares

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

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

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

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

have a nice day ! from the Philippines...