TITLE: Proving irreducibility of $x^6-72$
QUESTION [8 upvotes]: I have the following question:

Is there an easy way to prove that $x^6-72$ is irreducible over $\mathbb{Q}\ $?

I am trying to avoid reducing mod p and then having to calculate with some things like $(x^3+ax^2+bx+c)\cdot (x^3+dx^2+ex+f)$ and so on...
Thank you very much.

REPLY [2 votes]: We can, after suitable substitution, use criterion given by Osada in Theorem 2.2.7 in Prasolov's great book Polynomials: 

Let $f(x)=x^n+a_1x^{n-1}+\dots+a_{n-1}x\pm p$ be a polynomial with integer coefficients, where $p$ is a prime. If $p>1+|a_1|+\dots+|a_{n-1}|$, then $f$ is irreducible.

In this case we have
$$f(x+1)=x^6+6x^5+15x^4+20x^3+15x^2+6x-71$$ 
which satisfies the criterion since $71$ is a prime and $71>1+6+15+20+5+6 =53$, hence the $f(x)$ is irreducible.