Prove that $\sqrt[n]{2}+\sqrt[n]{3}$ is irrational for every natural $n \ge 2$

Question

I want to prove that that $\sqrt[n]{2}$ + $\sqrt[n]{3}$ is irrational for every natural $n \ge 2$.

I tried to use some theorem of minimal polynomials, but I get nothing. Also i tried to assume that this number is rational $r=\sqrt[n]{2}+\sqrt[n]{3}$ and take $3 = (r-\sqrt[n]{2})^n = ...$. I think that this will be a good way, but we have here a little mess.

Answer 1

The polynomial $P(z) = z^n - 2$ is irreducible over the rationals by Eisenstein's criterion. So if $\alpha$ is a root of $P(z)$, $P(z)$ is its minimal polynomial over the rationals. Now if $(\alpha - r)^n = b$ is rational where $r$ is a nonzero rational, then $Q(z) = (z - r)^n - b - (z^n - 2)$ is a nontrivial polynomial of degree $n-1$ over the rationals such that $Q(\alpha) = 0$.

Answer 2

To add to Bob Israel's good answer, there is a general pattern of proving that what strikes us as plausible is true: nth roots of (for example) distinct primes are as algebraically independent as we imagine they could be. There have been several questions on this site about it, with many good answers, and I am too lazy to log them. I am just barely not-lazy enough to link to a PDF I wrote up about this kind of issue a while back, at http://www.math.umn.edu/~garrett/m/v/linear_indep_roots.pdf