How do I factorize x^{6}-1 over GF(3)? I know that

Step 1
If ${\displaystyle n=\pm }$ is a multiple of p then over ${\displaystyle GF\left(p\right)}$ one has
${\displaystyle x^n-1={(x^m-1)}^p}$
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)
${\displaystyle x^n-1=\prod _{k=0}^{n-1}(x-{\zeta }^j)}$
where ${\displaystyle \zeta }$ 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
${\displaystyle (x-{\zeta }^k)(x-{\zeta }^{pk})(x-{\zeta }^{p^{2}k})\cdots (x-{\zeta }^{p^{r-1}k})}$
has coefficients in, and is irreducible over, GF(p) where r is the least positive integer with ${\displaystyle p^{r}k\equiv k}$ (mod n)
Using this, it's easy to work out the degrees of the irreducible factors of ${\displaystyle x^n-1}$, but to find the factors themselves needs a bit more work, using for instance Berlekamp's algorithm.

Step 1
Factoring the polynomials ${\displaystyle x^n-1}$ 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.