How to find the splitting field and Galois group of $x^6 -4x^3 +1$? 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 $D_{12}$. 
 A: 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)
A: So reduction mod 11 shows there is an involution of the type $(\cdot\cdot)(\cdot\cdot)$, and reduction mod 5 shows there is a 6-cycle, in your Galois group.  Note also that replacing $x^3$ by $y$ shows your group has a normal subgroup of index $2$. In terms of intermediate fields, this comes from $\mathbb{Q}(\sqrt{3})$.
Now from the equation $x^3=2\pm\sqrt{3}$, we know also that  cubic root of unity $\dfrac{-1-i\sqrt{3}}{2}$ is in your splitting field. But your splitting field already contains $\sqrt{3}$, and that shows $i$ is in your splitting field. Finally, there is the element $\omega$, which is the real cubic root of $2+\sqrt{3}$.  Since the real cubic root of $2-\sqrt{3}$ can be written as $(2-\sqrt{3})\omega^2$, your splitting field is $\mathbb{Q}(\omega,i,\sqrt{3})$. [The fact it can't be any smaller comes from considering real subfields, and relatively prime degree extensions.]
So the Galois group stuff shows the extension is at least of degree $12$, and the explicit splitting field shows it is at most of degree $12$, so it is degree $12$.
You can now go on and show the Galois group is $D_{12}$ through whichever means you like the most. [For me, it would be by proving there is only one non-abelian group of order $12$ with a cyclic subgroup of order $6$, but no cyclic subgroup of order $4$.]
