How to prove that $\mathbb{Q} (\sqrt[3]{2}+\sqrt[3]{4})= \mathbb{Q} (\sqrt[3]{2})$.
"$\subset$" Since $\sqrt[3]{2} \in \mathbb{Q} (\sqrt[3]{2}) $ and $\sqrt[3]{4} \in \mathbb{Q} (\sqrt[3]{2}) $ (because $\sqrt[3]{4} = (\sqrt[3]{2})^2$), thus the sum $\sqrt[3]{2}+\sqrt[3]{4} \in \mathbb{Q} (\sqrt[3]{2})$
"$\supset$" I don't know. I know that $ \mathbb{Q} (\sqrt[3]{2}) = \{a+b\sqrt{2}+c \sqrt[3]{2}: \, a,b,c \in \mathbb{Q} \}$ but $\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{4}) =?$
To prove $F(\alpha)=F(\beta)$ for a field $F$, exactly as you used, we have to prove both containments, i.e. that $\alpha\in F(\beta)$ and $\beta\in F(\alpha)$, i.e. that they can be expressed by each other using field operations and scalars from $F$.
Let $\alpha:=\sqrt[3]2$ (so that we use $\alpha^3=2$), and let$\beta:=\sqrt[3]2+\sqrt[3]4=\alpha+\alpha^2$.
Then $\beta^2=\alpha^2+2\alpha^3+\alpha^4=\alpha^2+4+2\alpha$.
From this, we can express $\alpha$ as $\beta^2-\beta-4$.
