Prove that there exists a polynomial $p(x) \in \mathbb{Q}[x]$ of degree 21 such that $p(5^{1/3}+7^{1/7})=0$

Question

Prove that there exists a polynomial $p(x) \in \mathbb{Q}[x]$ of degree 21 such that $p(5^{1/3}+7^{1/7})=0$

It would be nice if I had some simple theorem to gaurentee that $\mathbb{Q}(5^{1/3}+7^{1/7})=\mathbb{Q}(5^{1/3},7^{1/7})$.

Other than that, I was going to set $u=(5^{1/3}+7^{1/7})$ and try to find a degree 21 polynomial and then prove it was irreducible. Anyone have any insight into this?

Answer 1

What we know for sure is that $5^{\frac{1}{3}}+7^{\frac{1}{7}}\in\mathbb{Q}(5^{\frac{1}{3}},7^{\frac{1}{7}})$ and hence $\mathbb{Q}(5^{\frac{1}{3}}+7^{\frac{1}{7}})\subseteq\mathbb{Q}(5^{\frac{1}{3}}, 7^{\frac{1}{7}})$. So the degree of the minimal polynomial of $5^{\frac{1}{3}}+7^{\frac{1}{7}}$ over $\mathbb{Q}$ divides $21$. Can you finish from here?