Can someone give me an idea how to prove :
a) that the minimal polynomial of $\sqrt[3]{2}\in F$ (where $F$ is the splitting field of $x^3-2$ over $\mathbb{Q} $) over the field $\mathbb{Q}(r\sqrt[3]{2})$ has degree 2 (hence the name of the title), where $r=e^{\frac{2\pi i}{3}}$ (notice that the roots of $x^3-2$ are $\sqrt[3]{2},r\sqrt[3]{2},r^2 \sqrt[3]{2}$);
b) that $\sqrt[3]{2}\not \in \mathbb{Q}(r\sqrt[3]{2})$ ?
What I know until now for a): I know that $\mathbb{Q}(r\sqrt[3]{2})(\sqrt[3]{2})=F$, since from $r\sqrt[3]{2},\sqrt[3]{2}$ I can recover $r^22^{\frac{1}{3}}$ (and by having all roots of the polynomial, the above must coincide with the splitting field); I think the polynomial I'm looking for is $x^2+\sqrt[3]{4}$, but I don't know how to prove that $\sqrt[3]{4} \in \mathbb{Q}(r\sqrt[3]{2})$, since this time I have no clue how to recover $\sqrt[3]{4}$ only from $r\sqrt[3]{2}$. For b) I think this must have something to do with the fact, that $\mathbb{Q}(r\sqrt[3]{2})=\{a+ b\cdot r\sqrt[3]{2}+ c \cdot r^2 \sqrt[3]{4}|a,b,c\in \mathbb{Q}\}$, since $(r\sqrt[3]{2})^3=1\cdot 2=2$, but I can't figure out how to turn this observation into a precise argument (because one could argue that it still might be possible to obtain $\sqrt[3]{2}$ if one only had suitable $a,b,c$)