How do you factorise the given polynomial x9-x6-x+1?

# How do you factorise the given polynomial ${x}^{9}-{x}^{6}-x+1$?

1. A
${\left(x-1\right)}^{2}{\left({x}^{2}+x+1\right)}^{2}\left(x+1\right)\left({x}^{2}-x+1\right)$
2. B
$\left({x}^{2}-1\right)$
3. C
$\left(x-1{\right)}^{2}{\left({x}^{2}+x+1\right)}^{2}$
4. D
$\left(x+1\right)\left({x}^{2}-x+1\right)$

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### Solution:

By grouping method
First factor two terms, factor ${x}^{6}$ and 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 to two square
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+\mathit{ab}+{b}^{2}\right)$
${a}^{3}{+b}^{3}=\left(a+b\right)\left({a}^{2}-\mathit{ab}+{b}^{2}\right)$
${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$
So that
$\left(x-1\right)\left({x}^{2}+x+1\right)\left({x}^{3}-1\right)\left({x}^{3}+1\right)$
$\left(x-1\right)\left({x}^{2}+x+1\right)\left(x-1\right)\left({x}^{2}+x+1\right)\left(x+1\right)\left({x}^{2}-x+1\right)$
$\left(x-1{\right)}^{2}{\left({x}^{2}+x+1\right)}^{2}\left(x+1\right)\left({x}^{2}-x+1\right)$ is the correct answer.

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