Showing $G(\mathbb{Q}(\sqrt[3]{2}, w)/ \mathbb{Q}) \simeq S_3$ I want to show that $G(\mathbb{Q}(\sqrt[3]{2}, w)/ \mathbb{Q}) \simeq S_3$, but I'm not that familiar with computing Galois groups so I don't really know how to do this exercise. 
How do I approach this?
Also note that we define $w$ as: $w = \frac{1}{2}(1 + i\sqrt{3})$ 

The extension $K/Q$ is Galois as its the splitting field of a seperable polynomial. The Galois group has prder 6. Notice that $\mathbf{Q}(\omega \sqrt[3]{2}$ and $\mathbf{Q}(\sqrt[3]{2})$ are both subfields of degree three! So you have at least two subgroups of order 2. Hence the Galois group is not cyclic, hence it's $S_{3}$, $K=\mathbf{Q}(\sqrt[3]{2},\omega)$. (I am using the fact that any group of order 6 is isomorphic to $\mathbf{Z}/6\mathbf{Z}$ or $S_{3}$ and that the Galois correspondence theorem!)
