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

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

Here is an answer which uses some elementary properties of field automorphisms (essentially the statement that degree $2$ extension is Galois, but this is something which can be rather easily proven directly using just field theory).

Let $K=\mathbb Q(\sqrt[p]{2}+\sqrt{5})$. Suppose $\sqrt{5}\not\in K$. Then $K(\sqrt{5})$ is a degree $2$ extension of $K$, which implies there exists an automorphism $\sigma$ of $K(\sqrt{5})/K$ such that $\sigma(\sqrt{5})=-\sqrt{5}$. Since $\sqrt[p]{2}+\sqrt{5}\in K$, it is fixed by $\sigma$, so we have $$\sigma(\sqrt[p]{2})=\sigma(\sqrt[p]{2}+\sqrt{5}-\sqrt{5})=\sigma(\sqrt[p]{2}+\sqrt{5})-\sigma(\sqrt{5})=\sqrt[p]{2}+\sqrt{5}+\sqrt{5}.$$

However, $\sqrt[p]{2}+2\sqrt{5}$ is not a conjugate of $\sqrt[p]{2}$. Indeed, since $p>2$, the nontrivial conjugates of $\sqrt[p]{2}$ are not real numbers, and clearly $\sqrt[p]{2}+2\sqrt{5}\neq\sqrt[p]{2}$. This is a contradiction, so we conclude $\sqrt{5}\in K$ after all.