Define by $\mathbb{Q}(\alpha,\omega)$ to be the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ along with elements $\alpha,\omega$. Now let $\alpha=\sqrt[6]{5}$ and $\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Prove that $\mathbb{Q}(\alpha,\omega),\mathbb{Q}(\alpha\omega^2,\omega^5),\mathbb{Q}(\alpha\omega^4,\omega^5)$ are in fact the same field.
Attempt: Since $\alpha,\omega\in\mathbb{Q}(\alpha,\omega)$, we have that any product is also in this field. For instance, $\omega^2\in\mathbb{Q}(\alpha,\omega)$ and $\alpha\omega^2\in\mathbb{Q}(\alpha,\omega)$. Similarly $\omega^5\in\mathbb{Q}(\alpha,\omega)$. So the first field is contained in the second. We now need to show the reverse, that $\mathbb{Q}(\alpha\omega^2,\omega^5)\subseteq\mathbb{Q}(\alpha,\omega)$.
I have no idea how to do this. In fact I am not even sure my first set of working is correct. How do we do this?