Degree of splitting field of $x^3-5$ over $\mathbb{F}_7$

Question

Find the degree of the splitting field of $f(x):=x^3-5$ over $F:=\mathbb{F}_7$.

Attempt:

$f$ is irreduicible in $F[x]$ (suppose in contradiction it is reducible, thus it splits to at least one linear normalized polynomial element but this is contradiction beacuse $f$ has no roots in $F$). Thus, if $E$ is the splitting field of $f$ over $F$, we get $F\subset F(5^{1\over3})\subseteq E$. So I get that the degree is at least $2$.

Answer 1

If $[E:F]=2$ and $\alpha^3=5$ then there exists some polynomial $g$ of degree $2$ such that $g(\alpha)=0$. But then $$f(x)=g(x)q(x)+r(x)$$ for some polynomials $q$ and $r$, and the degree of $r$ is at most $1$. Therefore $$0=f(\alpha)=q(\alpha)q(\alpha)+r(\alpha)=r(\alpha)$$ So $r=0$. This is a contradiction because $f$ is irreducible.

Answer 2

Hint: $f(x)=x^3-5$ is irreducible over $\Bbb F_7$ since it has no root.

Further reference:

Splitting field of $x^3 - 2$ over $\mathbb{F}_5$

Answer 3

Hint:

If $\omega$ is a root of $x^3-5$ in some extension (it has no root in $\mathbf F_7$), search for cube roots in $\mathbf F_7$. Deduce the splitting field is $\mathbf F_7[\omega]$.