Determine the degree of the field extension $\mathbb{Q}(\sqrt{2}+(5)^{1/3})$ over $\mathbb{Q}$.

Question

Determine the degree of the field extension $\mathbb{Q}(\sqrt{2}+(5)^{1/3})$ over $\mathbb{Q}$.

My idea to solve this question is, firstly I need to show $\mathbb{Q}(\sqrt{2}+(5)^{1/3})=\mathbb{Q}(\sqrt{2},(5)^{1/3})$. I calculated the degree and my answer is $6$.

I tried to prove $\mathbb{Q}(\sqrt{2}+(5)^{1/3}) = \mathbb{Q}(\sqrt{2},(5)^{1/3})$. But due to lengthy computations, I am unable to prove this subpart.

Anyone can suggest some hint to solve this subpart of the question?

Answer 1

Hint: if you already know that $\mathbb{Q} \big( \sqrt{2}, \sqrt[3]{5} \big)$ has degree $6$ over $\mathbb{Q}$, you may assume for contradiction that $\mathbb{Q} \big( \sqrt{2} + \sqrt[3]{5} \big) \subsetneq \mathbb{Q} \big( \sqrt{2}, \sqrt[3]{5} \big)$ and use the formula $[M:K] = [M:L] \cdot [L:K]$.