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 

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 $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.
