Degree of the extension $\mathbb{Q}(\sqrt[5]{7}+\sqrt[5]{49})$

Question

The original question is find the degree of the irreducible polynomial of $3+\sqrt[5]{7}+\sqrt[5]{49}$, but it's equivalent to find $[\mathbb{Q}(3+\sqrt[5]{7}+\sqrt[5]{49}):\mathbb{Q}]$.

$\mathbb{Q}(3+\sqrt[5]{7}+\sqrt[5]{49})=\mathbb{Q}(\sqrt[5]{7}+\sqrt[5]{49})=\mathbb{Q}(\sqrt[5]{7}+(\sqrt[5]{7})^{2})$. But i don't know how to continue. If the question was $\mathbb{Q}(\sqrt[3]{7}+(\sqrt[3]{7})^{2})$: $$\alpha=\sqrt[3]{7}+(\sqrt[3]{7})^{2}$$ $$\alpha^{3}=7+3*7\sqrt[3]{7}+3*7(\sqrt[3]{7})^{2}+49$$ $$\alpha^{3}=21\alpha+56$$ $$X^{3}-21X-56$$ Irreducible, because Eisenstein's criterion with $p=7$. But this method doesn't work with $n=5$, or at least it isn't clear to find the relation. I think there should be another method to solve this more easily.

Answer 1

Since $\mathbb Q\subseteq \mathbb{Q}(\sqrt[5]{7}+\sqrt[5]{49}) \subseteq \mathbb{Q}(\sqrt[5]{7})$ and $ [\mathbb{Q}(\sqrt[5]{7}):\mathbb Q]=5 $, we must $[\mathbb{Q}(\sqrt[5]{7}+\sqrt[5]{49}):\mathbb Q] =1 $ or $5$. But $\sqrt[5]{7}+\sqrt[5]{49}$ is not rational, for otherwise $\sqrt[5]7$ will be the root of some polynomial of the form $x^2+x-q$, where $q\in\mathbb Q$, contradicting the fact that $x^5-7$ is irreducible.

Answer 2

Hint. Let us call your element $\alpha$. It is in the field $\mathbf{Q}(\sqrt[5]{7})$, which is of degree $5$ over $\mathbf{Q}$. As $5$ is prime, it suffices by the tower law to prove that $\alpha\not\in \mathbf{Q}$ to conclude that $\alpha$ also has degree $5$ over $\mathbf{Q}$.