Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors.
This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. Suppose char $F\neq 2,3$. Suppose also that $f$ is neither irreducible nor splits in $F$. Then $f$ is reducible which implies that $a_1 \in F$. i.e. $f(x)=(x-a_1)g(x)$, where $g(x)\in F[x]$ is irreducible with deg $g=2$. Let $K$ be the splitting field of $g$. The $K$ is Galois over $F$.
So if $\sigma \in $ Aut($K/F$), then $\sigma (a_1)=a_1$ since $\sigma $ fixes $F$ and $a_1 \in F$. Since $\sigma$ permutes the roots of $f$, WLOG suppose $\sigma (a_2)=a_3$. Then
$\sigma(\triangle) = \sigma((a_1-a_2)(a_1-a_3)(a_2-a_3))=-\triangle$.
But $\triangle^2=D(f)=81$, so $\triangle = \pm 9 \in F$, so $\triangle \in F$. Therefore, $9=-9$ $\implies 1=-1 \implies$ char $F =2$ which is a contradiction. So $f$ is either irreducible or splits in $F$.
Next suppose char $F=2$. Well, I'm not exactly sure what I can say about $f$.
I would like to know if my approach is correct and also what to do in the second case. Thanks.
If char $F=2$, then $f=x^3+x+1$. Suppose $b$ is a root of $f$. Then $b^2 $ is also a root, since $f(b^2)=(b^2)^3+b^2+1=(b+1)^2+b^2+1=2b^2+2=0$. Is it enough to conclude that $f$ splits?