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 
 A: 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:
$\sqrt{2} = (\frac{a}{2})^{-3}$
$\sqrt[3]{2} = (\frac{a}{2})^{-2}$.
A: 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 $\mathbb Q(\sqrt2)$. 
A: Try to express both $\sqrt{2}$ and $\sqrt[3]{2}$ as rational functions of $a = \sqrt{2}+\sqrt[3]{2}$. The job is simple and easily done via equation $$(a -\sqrt{2})^{3}=2\tag{1}$$ so that $$a^{3}-3\sqrt{2}a^{2}+6a-2\sqrt{2}=2$$ or $$\sqrt{2}=\frac{a^{3}+6a-2}{3a^{2}+2}\tag{2}$$ and we have $$\sqrt[3]{2}=a-\sqrt{2}$$ and using equation $(2)$ we can replace $\sqrt{2}$ by a rational function of $a$, so that $\sqrt[3]{2}$ is also a rational function if $a$. It thus follows that $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.
A: $\mathbb{Q}(\alpha,\beta)$ 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 $s = 1.$
