Let $x = e^{2\pi i/3} = -\frac{1}{2} + \frac{1}{2}\sqrt{3}i$. I want to find an element $\alpha \in \mathbb{Q}(x, \sqrt[3]{2})$ so that $\mathbb{Q}(x, \sqrt[3]{2}) = \mathbb{Q}(\alpha)$. I am not sure how to do this. Any help?
My gut tells me it may by $\alpha = x + \sqrt[3]{2}$, but I'm unsure of that.
As it follows from Dave's comment the equality $\mathbb{Q}(x,\sqrt[3]{2})=\mathbb{Q}(\sqrt{3}i,\sqrt[3]{2})$ holds. If you know that then it is not too hard to guess the number $\alpha=i\sqrt{3}\sqrt[3]{2}$. Any field which contains the elements $\sqrt{3}i$ and $\sqrt[3]{2}$ obviously contains $\alpha$ as well. So $\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\sqrt{3}i,\sqrt[3]{2})$. Also, if a field contains $\alpha$ then it contains $\alpha^3,\alpha^4$ as well, and as you can easily check it implies that the field contains $\sqrt{3}i$ and $\sqrt[3]{2}$. So $\mathbb{Q}(\sqrt{3}i,\sqrt[3]{2})\subseteq\mathbb{Q}(\alpha)$.