This exercise is taken from Falko Lorenz's Algebra 5.2.
Why is $\sqrt[3]{2}$ not an element of $\mathbb{Q}(\sqrt[7]{5})$? Why is there no extension E of $\mathbb{R}$ such that $E:\mathbb{R}=3$.
My Progress: The first question I have answered as follows (of which I am not certain of my answer). Could anyone give me a hint on the second question?
Suppose $\sqrt[3]{2}$ is an element of $\mathbb{Q}(\sqrt[7] {5})$. Then $\mathbb{Q}(\sqrt[3]{2})\subset \mathbb{Q}(\sqrt[7]{5})$. This implies that $[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}]$ But since $X^3-2$ is irreducible in $\mathbb{Q}[X]$ with root $\sqrt[3]{2}$, $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and similarly $X^7-5$ is irreducible in $\mathbb{Q}[X]$ with root $\sqrt[7]{5}$, $[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}]=7$, contradiction.