TITLE: Prove that $\sqrt[3]{p}$, $\sqrt[3]{q}$ and $\sqrt[3]{r}$ cannot be in the same arithmetic progression
QUESTION [6 upvotes]: My cousin (he doesn't speak English well so I am writing on his behalf) is trying to do the following problem:
Let $p$,$q$, $r$ be different primes (let's assume $p<q<r$). Show that $\sqrt[3]{p}$, $\sqrt[3]{q}$ and $\sqrt[3]{r}$ cannot be in the same arithmetic progression. 
I honestly don't know how to do it without going into field extensions and that kind of stuff, but my cousin is at high school level (he's preparing for IMO or something similar). So here's what we have done so far:
Suppose there exists such a progression: $a[n] = nd + a[0]$, where $a[n_0] =\sqrt[3]{q}$ and $ a[n_1] =\sqrt[3]{r}$
WLOG assume that $\sqrt[3]{p} = a[0]$
Then it must be $$\frac{\sqrt[3]{q}-\sqrt[3]{p}}{n_0} = \frac{\sqrt[3]{r}-\sqrt[3]{p}}{n_1} $$
Therefore $\frac{\sqrt[3]{r}-\sqrt[3]{p}}{\sqrt[3]{q}-\sqrt[3]{p}}\in \mathbb{Q}$. We want to show that this can't be, but alas we haven't succeeded. We have tried to exponentiate, take the roots out of the denominator, etc. with no success. Any help will be most appreciated. Thanks!
Note: We have to prove that they are not in the same arithmetic progression, consecutive or not (arbitrary positions)

REPLY [2 votes]: If $a(\sqrt[3]r - \sqrt[3]p)=b(\sqrt[3]q - \sqrt[3]p)$ with $a,b \in \Bbb Z$ (with $a,b$ nonzero and distinct),
then $(b-a)\sqrt[3]p = b\sqrt[3]q - a\sqrt[3]r$,
and $(b-a)^3p = b^3q - 3ab\sqrt[3]{qr}(b\sqrt[3]q-a\sqrt[3]r)-a^3r = b^3q-3ab(b-a)\sqrt[3]{pqr} - a^3r$.
Then, $\sqrt[3]{pqr} = (a^3r-b^3q+(b-a)^3p)/3ab(a-b)$ is rational. 
You can quickly deduce a contradiction from there.