Over the complex numbers this factors as .
Thus, the splitting field is given by .
Now, since .
The polynomial is irreducible over so it must be irreducible over because and . This means that .
And of course, .
The Galois group therefore must have order and that already gives us a few ideas of this group might be.
There are five possibilites: .
I will let you think about which one out of the five it must be. But before you do that it would be helpful to make the following realizations. First, if is an automorphism then it is completely determined by the values . Now, must permute the roots of the minimal polynomial of each element, so, . Thus, there are a total of at most such permutations but because the Galois group has order twelve it means for each one of these possibilities we have a permuation. Let be the permutation: i.e. is complex conjugation. Let be the permutation: . Let be the permutation: . Notice that are non-trivial automorphisms with .
Remember that , so, . While . We see that so the Galois group is not abelian. This eliminates the first two abelian cases and leaves us with only . Can you argue from here?