Factorize $x^9 - 1 $ in $\mathbb F_3[x]$ 
I was able to factorize $x^9-x$ in $\mathbb F_3[x]$, however, with $x^9-1$ I am not sure what to do. 

My first intuition was to do $(x^3-1)(x^3+1)$ but this is clearly just the first step, if not completely misplaced.
Thanks for help!
 A: By Fermat's little theorem, over $\mathbb{F}_3$ we have
$$ x^9-1 \equiv (x-1)^9.$$
A: $x\mapsto x^3$ is a field automorphism in characteristic $3$, hence
$$x^9-1=x^{3^2}-1^{3^2}=(x-1)^{3^2}=(x-1)^9.$$
A: Note that 
\begin{eqnarray*}
(x+2)^9=x^9+18x^8+144x^7+672x^6+2016x^5+4032x^4+5376x^3+4608x^2+2304x+512
\end{eqnarray*}
Now notice that all of these coefficients are divisible by $3$ apart from the first and the last.
So $x^9-1=(x+2)^9$ modulo 3.
A: This is a very "hands-on" answer.
There are only three elements of $\mathbb F_3$, i.e. $0$, $1$ and $2$.
The remainder theorem tells us that if $\mathrm f(a)=0$ then $(x-a)$ is a factor of $\mathrm f(x)$.
Let's try $x=0,1,2$ and see what happens:
\begin{eqnarray*}
0^9-1 &\equiv& 2 \pmod 3 \\
1^9-1&\equiv& 0 \pmod 3 \\
2^9-1 &\equiv& 1 \pmod 3
\end{eqnarray*}
Since $\mathrm f(1) \equiv 0 \pmod 3$, it follows that $(x-1)$ is a factor of $\mathrm f(x)$.
You need to use polynomial division, and reduce modulo $3$.
$$\frac{x^9-1}{x-1} = 1+x+x^2+\cdots +x^7+x^8$$
Next, we need to use the Factor Theorem on $\mathrm{g}(x):=1+x+x^2+\cdots +x^7+x^8$.
\begin{eqnarray*}
\mathrm g(0) &\equiv& 1 \pmod 3 \\
\mathrm g(1) &\equiv& 0 \pmod 3 \\
\mathrm g(2) &\equiv& 1 \pmod 3
\end{eqnarray*}
It follows that $(x-1)$ divides $\mathrm g(x)$ modulo $3$. Using Polynomial division
\begin{eqnarray*}
\frac{\mathrm g(x)}{x-1} &=&
 x^7+2x^6+3x^5+4x^4+5x^3+6x^2+7x+8+\frac{9}{x-1} \\ \\
&\equiv& x^7+2x^6+x^4+2x^3+x+2 \pmod 3
\end{eqnarray*}
Let $\mathrm h(x) := x^7+2x^6+x^4+2x^3+x+2$ and proceed as before. 
We see that $\mathrm h(x) \equiv 0 \pmod 3 \iff x \equiv 1 \pmod 3$, so consider
\begin{eqnarray*}
\frac{\mathrm h(x)}{x-1} &=& x^6+3x^5+3x^4+4x^3+6x^2+6x+7+\frac{9}{x-1} \\ \\
&\equiv& x^6+x^3+1 \pmod 3
\end{eqnarray*}
Again, we see that $(x-1)$ divides $x^6+x^3+1 \pmod 3$, leaving $x^5+x^4+x^3+2x^2+2x+2$, which is divisible by $x-1$. This leaves $x^4+2x^3+2x+1$, which is divisible by $x-1$. This gives $x^3+2$. This is divisible by $x-1$ which leaves $x^2+x+1$. This is divisible by $x-1$, and that leaves $x+2$. This is congruent to $x-1$.
\begin{eqnarray*}
(x-1)^9 &=& x^9-9x^8+36x^7-84x^6+126x^5-126x^4+84x^3-36x^2+9x-1 \\
&\equiv& x^9 + 2 \pmod 3 \\ 
&\equiv& x^9 -1 \pmod 3
\end{eqnarray*}
A: Recall: $ $ Frobenius / Freshman's Dream: $\ \ (x+y)^{\large p} =\, x^{\large p} + y^{\large p}\, $ in $\,\Bbb F_p.\,$ Applied twice we have
$$\begin{align} (x-y)^{\large p} &= \ x^{\large p} - y^{\large p}\\
\overset {(\ \  \ )^{\Large p}}\Longrightarrow\ (x-y)^{\large p^{\Large 2}}\! &= x^{\large p^{\Large 2}}\!\! - y^{\large p^{\Large 2}} \end{align}$$
