Factorize $x^3-1$ in $F_3[x]$ into irreducible polynomials

Question

I'm fairly new to this topic:

We can see that $x^3-1=(x-1)(x^2+x+1)$

But what I don't understand is how $(x^2+x+1)$ can be further reduced to $(x-1)^2$

Many thanks

Answer 1

Here $(x-1)^2=x^2-2x+1=x^2+x+1$ since $-2=1$ in $\Bbb F_3$.

(More generally, the Frobenius isomorphism says $(x+y)^p=x^p+y^p\pmod p$ for any prime $p$. Or, more generally, the map $a\mapsto a^p$ is an isomorphism in $\Bbb F_p[x]$.)