TITLE: Showing that $\sqrt{5} \in \mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$
QUESTION [8 upvotes]: I am attempting to show that $\sqrt{5} \in \mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$, where $p > 2$ is prime. I have already shown that $[\mathbb{Q}(\sqrt[p]{2}, \sqrt{5}) : \mathbb{Q}] = 2p$.
If needs be, I can understand that this might constitute proving that $\mathbb{Q}(\sqrt[p]{2} + \sqrt{5}) = \mathbb{Q}(\sqrt[p]{2}, \sqrt{5})$, which I know intuitively but am unsure how to prove. In that regard, I am aware of questions such as this and this, but all answers provided either

do not seem to generalize easily to cases where not both of the roots are square.
are beyond the scope of my current course.

Any help towards a (preferably low-level) proof of either the inclusion of $\sqrt{5}$ or the equality of $\mathbb{Q}(\sqrt[p]{2} + \sqrt{5})$ and $\mathbb{Q}(\sqrt[p]{2}, \sqrt{5})$ is much appreciated.

REPLY [7 votes]: So let $\alpha = \sqrt [p]2+\sqrt 5$ then $$\left(\alpha -\sqrt 5\right)^p=2$$
Expand the left-hand side using the binomial theorem to obtain $$p(\alpha)-q(\alpha)\sqrt 5=2$$ where $q(\alpha)\gt 0$ since all the terms involving $\sqrt 5$ have the same sign. Also $p(\alpha), q(\alpha)$ are polynomials in $\alpha$ and belong to $\mathbb Q(\alpha)$
 Finally $$\sqrt 5=\frac {p(\alpha)-2}{q(\alpha)}$$