I want to find generators of the field extension $\mathbb{Q}(i,\sqrt[3]{2})$ over $\mathbb{Q}$. Using the primitive element theorem (see here), I get $pi+q\sqrt[3]{2}$, for every nonzero rational $p,q$, is a generator for the field extension (i.e. $\mathbb{Q}(i,\sqrt[3]{2})=\mathbb{Q}(\gamma)$ where $\gamma=pi+q\sqrt[3]{2}$ for any nonzero rational $p,q$).
My question is, are these the only possibilities for the generator, or can there be a generator of different form? How can I show this?