Is $\mathbb{Q}(\sqrt{3},\sqrt[3]{3})$ a splitting field over Q?

Question

Im having trouble showing that $\mathbb{Q}(\sqrt{3},\sqrt[3]{3})$ is a splitting field over $\mathbb{Q}$.
Im pretty sure it is, since $Irr(\sqrt[3]{3},\mathbb{Q})=x^3-3$ ,which has 2 complex roots. That shows that $\mathbb{Q}(\sqrt[3]{3})$ cannot be a splitting field over $\mathbb{Q}$. If we wanted to normalize $\mathbb{Q}(\sqrt[3]{3})$ ,we would extend to $\mathbb{Q}(\sqrt[3]{3},i)$.
Any tips on how to proceed?