Show that $\mathbb{Q}(\sqrt2, \sqrt[3]2)$ is a primitive field extension of $\mathbb{Q}$.

I've tried a method similar to showing that $\mathbb{Q}(\sqrt2, \sqrt3)$ is a primitive field extension, but the cube root of 2 just makes it a nightmare.

Thanks in advance

I think there is a simpler solution.

Let $a = \sqrt{2} \sqrt[3]{2}$.

$a$ is clearly in the field extension and both generators can be generated by it:

$\sqrt{2} = (\frac{a}{2})^{-3}$

$\sqrt[3]{2} = (\frac{a}{2})^{-2}$.

Let $\alpha=\sqrt2+\sqrt[3]2$. It's clear that $\mathbb Q(\alpha)$ is a subextension of $\mathbb Q(\sqrt2,\sqrt[3]2)$. All that remains is to show that $\mathbb Q(\alpha)$ has degree $6$ over $\mathbb Q$.

You could do this by explicitly calculating the minimal polynomial of $\alpha$ over $\mathbb Q$, or by observing that $$(\alpha-\sqrt2)^3=2,$$ which can be used to deduce that $\mathbb Q(\alpha)$ is a degree $3$ extension of $\mathbb Q(\sqrt2)$.