Factorize $x^8+2$ over $\mathbb{F}_3$ using cyclotomic cosets 
Determine the splitting field of $x^8 + 2$ over $\mathbb{F}_3$ and factor it into irreducible polynomials over $\mathbb{F}_3$ using cyclotomic cosets.

I'm having issues in factorizing this polynomial. can anyone see if I am worng with the computations or something else please?
My attempt: By a Lemma we had in the lecture:

Lemma: The splitting field $\mathbb{F}_{q^s}$ $\text{split}(f,\mathbb{F}_q)$, of $f=x^n-1$, is characterized through the smallest integer $s$ such that $n|q^s-1$.

Notice that over $\mathbb{F}_3$ we have $x^8+2\equiv x^8-1$.
For $n=8$ and $q=3$, $n|3^2-1=8\Rightarrow \text{split}(x^8+2,\mathbb{F}_3)=\mathbb{F}_{3^2}=\mathbb{F}_9$.
Consider that 2 is a root of $x^8-1$, i.e. $2^8-1=255\equiv 0\mod3$
Then I find the cyclotomic cosets $C_0=\{0\}$, $C_1=\{1,3\}$, $C_2=\{2,6\}$, $C_3=\{1,3\}=C_1$, $C_4=\{4\}$, $C_5=\{5,7\}$
So the factorization gives: $x^8-1=(x-2^0)\cdot(x-2^1)(x-2^3)\cdot(x-2^2)(x-2^6)\cdot(x-2^4)\cdot(x-2^5)(x-2^7)\equiv(x+2)(x^2+2x+1)(x^2+x+2)(x+2)(x^2+2x+1)$
But checking the L.H.S., this doesn't give $x^8-1$...

Here $x^8+2 = x^8-1 = (x^4-1)(x^4+1)$ with
$x^4+1 = (x^2-1)(x^2+1)$ and $x^2-1 = (x-1)(x+1)$.
Now $x^2+1 = (x+i)(x-i)$ where $i$ is a primitive 4th root of unity in an extension field of $F_3$.
Here $2^2 = 4 = 1$ in $F_3$ and so 2 is a primitive 2nd root of unity. Thus $x^2+1$ does not factor over $F-3$.
Similarly, $x^4+1$ does not factor over $F_3$ as $F_3$ does not contain a primitive 8th root of unity.
This gives the overall decomposition.
