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 

$\mathbb{Q}(\alpha,\beta)$ will equal $\mathbb{Q}(\alpha + s\beta)$ if you choose any rational number $s$ that is not $-\frac{\alpha_i - \alpha}{\beta_j - \beta}$ for any of the conjugates $\alpha_i$ of $\alpha$ and $\beta_j$ of $\beta.$
In your case you can take $s = 1.$
