Find the Galois group of the polynomial $f(x) = x^3 -10$ over the field $\mathbb{Q}(\sqrt{-3})$, given that the discriminant of $f$ is $-2700.$
I know that the Galois group of a given polynomial is contained in the permutation group on the roots of a given separable polynomial up to isomorphism. Using this theorem I have worked out the following:
$f(x) = x^3 - 10 \in \mathbb{Q}(\sqrt{-3})$
$\mathbb{Q}(10^\frac{1}{3},\sqrt{-3})$ is the minimal splitting field of $f(x)$ over $\mathbb{Q}(\sqrt{-3})$
Also, $[\mathbb{Q}(10^\frac{1}{3},\sqrt{-3}):\mathbb{Q}(\sqrt{-3})] = 3$
The roots $f$ are $\{10^\frac{1}{3}, 10^\frac{1}{3} \omega, 10^\frac{1}{3} \omega ^2\}$
Hence the Galois group of $f(x)$ is contained in $S_3(S)$ up to isomorphism and $O(S_3(S))=6$
The question had mentioned value of discriminant of $f$ but I haven't used it above. Am I missing something? Any help would be appreciated.
Edit: I have come across another result that
The discriminant of f is a square in k iff the group of f over k consists of even permutations only.
Using the above result can we conclude that Galois group of the polynomial f(x) will consist of only even permutation.