# How do I factorize $$\displaystyle{x}^{{{6}}}-{1}$$ over GF(3)?

How do I factorize ${x}^{6}-1$ over GF(3)? I know that the result is ${\left(x+1\right)}^{3}{\left(x+2\right)}^{3}$, but I'm unable to compute it myself.
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Pubephenedsjq
Step 1
If $n=±$ is a multiple of p then over $GF\left(p\right)$ one has
${x}^{n}-1={\left({x}^{m}-1\right)}^{p}$
so the problem reduces swiftly to the case where n is co' to p.
If p is not a factor of n then over an algebraic closure of GF(p)
${x}^{n}-1=\prod _{k=0}^{n-1}\left(x-{\zeta }^{j}\right)$
where $\zeta$ is a primitive n-th root of unity. One makes this into a factorization over GF(p) by combining conjugate factors together. For each k, the polynomial
$\left(x-{\zeta }^{k}\right)\left(x-{\zeta }^{pk}\right)\left(x-{\zeta }^{{p}^{2}k}\right)\cdots \left(x-{\zeta }^{{p}^{r-1}k}\right)$
has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with ${p}^{r}k\equiv k$ (mod n)
Using this, it's easy to work out the degrees of the irreducible factors of ${x}^{n}-1$, but to find the factors themselves needs a bit more work, using for instance Berlekamp's algorithm.
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Matronola3zw6
Step 1
Factoring the polynomials ${x}^{n}-1$ is very different from factoring general polynomials, since you already know what the roots are; they are, in a suitably large finite extension, precisely the elements of order dividing n. Since you know that the multiplicative group of a finite field is cyclic, the conclusion follows from here.