Let $f(x) = x^3-5$. Let $L/\mathbb{Q}$ be the splitting field of $f$ over $\mathbb{Q}$. Compute $[L:\mathbb{Q}]$ and find the Galois group $G = \text{Aut}_{\mathbb{Q}}(L)$.
My current thought is since every $L/\mathbb{Q}$ is separable and since $L/\mathbb{Q}$ is the splitting field of $f$ over $\mathbb{Q}$, $L/\mathbb{Q}$ is Galois. By solving the roots of $f(x)$, I can see $L=\mathbb{Q}(\omega, \sqrt{3}[5])$. However, I don't know how to proceed from here.
What are the degrees of $\mathbb Q(\omega)$ and $\mathbb Q(\sqrt[3]{5})$ over $\mathbb Q$? Do you see how this gives you $[L:\mathbb Q]$?
The Galois group $G$ is a transitive subgroup of $S_3$, so its order is either 3 or 6. But $f$ has a two complex roots, so complex conjugation yields an automorphism of order 2. Thus $G = S_3$.