Splitting field for the polynomial $\ x^3-2$ over $\mathbb Q$ is $\mathbb Q(\sqrt[3]{2},\sqrt{-3})$. Now roots of the above polynomial are $\sqrt[3]{2}, \sqrt[3]{2}\omega, \sqrt[3]{2}{\omega}^2$ Since they are the roots of polynomial we have $\mathbb Q(\sqrt[3]{2})$ isomorphic to $\mathbb Q(\sqrt[3]{2}\omega)$ as vector space over $\mathbb Q$. This implies that degree of $\mathbb Q(\sqrt[3]{2},\sqrt{-3})$ over $\mathbb Q(\sqrt[3]{2}\omega)$ is 2. Then what will be its basis which should contain 2 elements? Also, does any of the $\sqrt[3]{2}$ and $\sqrt{-3}$ belong to field $\mathbb Q(\sqrt[3]{2}\omega)$?
Edit: By contradiction I concluded that $\sqrt{-3} \notin \mathbb Q(\sqrt[3]{2}\omega)$.