Let $p,q,r$ be prime integers with $q\neq r$. Let $\sqrt[p]q$ denote any root of $x^p-q$ and $\sqrt[p]r$ denote any root of $x^p-r$. Please show that $\mathbb{Q}(\sqrt[p]q)\neq\mathbb{Q}(\sqrt[p]r)$.
By Eisenstein we know the two polynomials are irreducible and so they are minimal polynomials and both have degree $p$.
Edit: In my answer we require that $p,q,r$ be distinct prime numbers.
Consider first $K = \Bbb{Q}(\sqrt[p]{q})$. The polynomial $x^p - q$ as we know is irreducible by Eisenstein and so is the minimal polynomial of $\sqrt[p]{q}$. Now I calculate that $$\operatorname{disc} \left(\sqrt[p]{q}\right) = (-1)^{p(p-1)/2}p^p (-q)^{p-1}$$
and since $\operatorname{disc} \mathcal{O}_K$ divides this it follows that $r$ is unramified in $\mathcal{O}_K$. On the other hand if $L = \Bbb{Q}(\sqrt[p]{r})$ is equal to $K$ above then $\mathcal{O}_K = \mathcal{O}_L$ which contradicts the fact that $r$ totally ramifies in $\mathcal{O}_L$. The reason that $r$ totally ramifies in $\mathcal{O}_L$ is because the minimal polynomial of $\sqrt[p]{r}$ is Eisenstein at $r$.