$x^3-9$ is irreducible over $\mathbb{Z}$

Question

I am trying to prove that $x^3-9$ is irreducible over $\mathbb{Z}$.

The approach I usually try is Eisenstein's criterion, and the fact that $f\in\mathbb{Z}[X]$ irreducible $\iff$ $f(mx+n)\in\mathbb{Z}[X]$ irreducible. The thing is, every value I plug in seems to fail the condition "$p^2$ does not divide $a_0$". I looked at $f(x\pm1),f(x\pm 2),f(x\pm3)$.

The other thing I tried is $x^3-9$ will be irreducible if it is irreducible in $\mathbb{F}_p[X]$ for some $p$. Again, small values don't work. Apparently $p=31$ does the job, but that feels disproportionate.

Am I missing something?

Answer 1

It's much easier than that, as it's degree $3$, then it should have a linear factor (the only ways to factor it is as $3$ degree $1$ polynomials or $1$ degree $1$ and $1$ degree two), as it doesn't have a root in $\Bbb Z$ (use the rational root theorem), it's irreducible.

Answer 2

Since $x^3 - 9$ is a cubic, if it could be factored one of the factors would have to be linear. Furthermore, this factor would have to be of the form $(x-a)$, where $a$ is a divisor of $9$. You can then easily verify that none of the divisors of $9$ (positive or negative) are roots of the polynomial, and you're done.

Answer 3

If you really want to use $\mathbb{F}_p[X]$ for some prime $p$, note that the congruence $x^3\equiv 9\pmod{7}$ has no solutions.