abstract algebra - Showing that $sqrt{5} in mathbb{Q}(sqrt[p]{2} + sqrt{5})$

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.

Answer

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 $$\sqrt 5=\frac {p(\alpha)-2}{q(\alpha)}$$