I'm stuck between two possibilities for a Galois group of an extension, and I could use a hint on how to decide.
First we consider the Galois group of the splitting field of $h(X) = X^3 + 3X + \sqrt 3\in \mathbb Q(\sqrt3)[X]$ over $\mathbb Q(\sqrt 3)$ (that $h$ is prime and seperable is easy to see). Observe that if $\alpha$ is a root of $h$ then the minimal polynomial of $\alpha$ is $f(X) = X^6 - 6X^4 + 9X^2 - 3$, meaning that $[\mathbb Q(\alpha):\mathbb Q] = 6$. The discriminant of $h$ is $\Delta(h) = 27 = (3\sqrt3)^2$ i.e. it is a square in $\mathbb Q(\sqrt3)$. By multiplicativity of degree, this forces $\operatorname{Gal}(h/\mathbb Q(\sqrt 3))\simeq C_3$.
The last part in fact means that $\mathbb Q(\alpha)/\mathbb Q(\sqrt 3)$ is Galois, which in this case (not in general) implies that $\mathbb Q(\alpha)/\mathbb Q$ is Galois (in both cases the condition that needs checking is normality).
The goal is to calculate $\operatorname{Gal}(\mathbb Q(\alpha)/\mathbb Q)$.
A further observation is that the Galois group of the Galoisian closure of $\mathbb Q(\alpha^2)/\mathbb Q$ is also $C_3$, which, since we already know $[\mathbb Q(\alpha) : \mathbb Q] = 6$, means that $\operatorname{Gal}(\mathbb Q(\alpha)/\mathbb Q(\alpha^2))\simeq C_2$.
So there is a subgroup of order $2$, and a subgroup of order $3$. If there are more subgroups of order $2$, the group we're trying to calculate is $S_3$. If not, it is $C_6$. This is a practice exam question, which means there should be something I can do that isn't just loading up Maple and calculating roots (which I could do since $X^6 - 6X^4 + 9X^2 - 3 = (X^3 + 3X + \sqrt 3)(X^3 + 3X - \sqrt 3)$). However it isn't coming to me.
A hint mostly.
Note that if $a$ is a root of $$x^3+x+\sqrt{3}=0$$ then $-a$ is a root of $$x^3+x-\sqrt{3}=0$$
Thus permutations of the roots of these two polynomial are the same. Now use this to check if the Galois group is commutative.