How to compute a primitive element for the splitting field of $x^3-2 \in \Bbb{Q}[x]$? Let $\alpha:=\sqrt[3]{2}\in\mathbb{R}$ and $\omega:=e^{2\pi i/3}\in\mathbb{C}$. Then the splitting field for the polynomial $x^3-2\in\mathbb{Q}[x]$ is $$\mathbb{Q}(\alpha,\omega\alpha,\omega^2\alpha)=\mathbb{Q}(\alpha,\omega).$$
Since $\mathbb{Q}$ has characteristic zero we know from the Primitive Element Theorem that there exists some $\gamma\in\mathbb{Q}(\alpha,\omega)$ with $$\mathbb{Q}(\alpha,\omega)=\mathbb{Q}(\gamma).$$
Question: How can I find a specific example of such an element $\gamma$?


The primitive element theorem asserts that for $u \in K$ and $\alpha,\omega$ separables, $\alpha+\omega u$ is a primitive element of $K(\alpha,\omega)/K$ iff $\forall \sigma \in Gal(\overline{K}/K)$, $$\sigma(\alpha)+\sigma(\omega)u = \alpha +\omega u \implies \sigma(\alpha)=\alpha,\sigma(\omega) = \omega$$
only finitely many $u$ don't work because those not working are of the form $u=\frac{ \sigma(\alpha)-\alpha}{\omega - \sigma(\omega)}$.

Here $K=\Bbb{Q}, \alpha=  2^{1/3}, \omega = e^{2i \pi /3}$, from our knowledge of Galois groups $\sigma(\alpha) = \omega^l \alpha, \sigma(\omega) = \omega^m, 3 \nmid m$,
with $u=1$,
for $\sigma(\alpha)+\sigma(\omega)= \alpha +\omega $, if $l\ne 0$ then $\alpha= \frac{\omega^m-\omega}{1-\omega^l} \in \Bbb{Q}(\omega)$ a contradiction, thus $l = 0$ and $\sigma(\alpha) = \alpha,\sigma(\omega)=\omega$.
Whence $\Bbb{Q}(2^{1/3},e^{2i \pi /3})=\Bbb{Q}(2^{1/3}+e^{2i \pi /3})$.
