I am trying to find the splitting field $L$ of the $x^6 -4x^3 +1$ over $\mathbb{Q}$, and its Galois group.
Here are some things I have figured out. I did the usual trick of solving for $x^3 = 2\pm \sqrt{3}$, which shows that $i, \sqrt{3}$ are both in the Galois group.
I know that $\mathrm{Gal}(L/\mathbb{Q})$ is non-abelian and I tried reducing mod primes to see what cycles might appear. For example, $$x^6 -4x^3 +1 \equiv (x-6)(x^2+6x+3)(x-2)(x^2+2x+4)\bmod 11,$$ which shows that the Galois group has a $4$-cycle. But generally I am stuck beyond these things and it would be great if someone could at least tell me of a simple way of calculating the degree of the splitting field.
Edit: Thank you for your help Steve and Hurkyl! After reading your solutions, I think I figured out a concrete way to verifying that the Galois group is $D_{12}$ once we know that the group has order 12.
Let $\alpha = \sqrt[3]{2+\sqrt{3}}, \beta = \omega \sqrt[3]{2+\sqrt{3}},
\gamma = \omega^2 \sqrt[3]{2+\sqrt{3}}$.
We know that the roots come in pairs $\alpha, \alpha^{-1}, \beta, \beta^{-1}, \gamma, \gamma^{-1}$ and the Galois group takes pairs to pairs. Also,
$\alpha \beta \gamma =1$ and so the triple $\alpha, \beta, \gamma$ either goes to the same triple $\alpha, \beta, \gamma$ or to the triple $\alpha^{-1}, \beta^{-1}, \gamma^{-1}$. Now if we label the vertices of a hexagon with the roots so that the pairs of roots are opposite vertices, then we see that the Galois group must be a symmetry of the hexagon (here must use both the relations that the pairs of roots go to pairs and that triples go to triples). Thus we see that the Galois group is contained in $D_{12}$ under this identification. Since the order of the group is $12$, the group must be $D_{12}$.
Best Answer
You were almost finished with the calculation of $L$.
Let $\alpha = \sqrt[3]{2 + \sqrt{3}}$ and $\beta = \sqrt[3]{2 - \sqrt{3}}$.
You've already observed that $\mathbb{Q}(\alpha^3) = \mathbb{Q}(\sqrt{3})$. And since $\alpha$ is a cube root of somethign in $\mathbb{Q}(\alpha^3)$ (that is not already a cube), we must also have $\omega \in L$, and so $K = \mathbb{Q}(i,\omega) \subseteq L$ (where $\omega$ is a primitive cube root of unity)
Finally, $K(\alpha)/K$ is an abelian extension of degree 3, and the same goes for $K(\beta) / K$ -- the only question is whether $\beta \in K(\alpha)$.
Well, this can be answered by looking at $K^*$. In this particular case, $\alpha^3$ is actually a unit of $\mathcal{O}$, the ring of integers in $K$. Furthermore, we know its unit group is $\mathbb{Z} \times \mu$, where $\mu$ is the roots of unity. Since $\alpha^3$ is not a root of unity, it generates a subgroup of infinite order.
The same goes for $\beta^3$. And therefore $\alpha^3$ and $\beta^3$ have to be related -- and now something that we probably should have noticed right from the beginning becomes the obvious thing to check for: we actually have $\alpha^3 \beta^3 = 1$. In fact, $\alpha \beta = 1$ in $L$.
So now it's clear: $L = K(\alpha) = \mathbb{Q}(i, \omega, \sqrt[3]{2 + \sqrt{3}})$. The six roots of your polynomial are:
$$ \alpha^{i} \omega^j $$
where $i \in \{-1,1\}$ and $j \in \{0,1,2\}$.
As for the Galois group, we've seen that it has a quotient isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}/2$. It's also clear it has a copy of $S_3$ as a subgroup, as that's the Galois group of $L / \mathbb{Q}(\sqrt{3})$. If I was up to par on my group theory, that would probably be enough for me to identify the group. (peeking at the other answer shows my first wild guess, $D_6$, would have been correct)