Let $F = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$. Show that $F$ is not normal over $\mathbb{Q}$.
Wikipedia says that an extension $L$ over $K$ is normal if when one root of its irreducible polynomial is in $L$, all the other are OR there are no roots in $L$
In order to show that this is not normal, I just need to find an irreducible polynomial that has one root in $F$ and a root not in $F$. An irreducible polynomial over $\mathbb{Q}$ can be $x^3-2$, which has a root in $F$ which is $\sqrt[3]{2}$. If we divide this polynomial by $x-\sqrt[3]{2}$, we get $x^2+\sqrt[3]{2}+2^{2/3}$ which has complex roots.
So is this a way to show that this field is not normal? What is $\sqrt{2}$ doing there?
How to show that it is indeed normal? I cannot solve for all irreducible polynomials, so is there a technique?
You did it well. Since $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)\subset\mathbb R$ and since $x^3-2$ has one but not all roots in $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)$ (since the other roots are non-real), $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)$ is not normal.