I'm trying to determine whether or not $x^6 + 108$ is irreducible over $\mathbb{Q}$. Is there an easy way to determine this ? I tried Eisenstein's Criterion directly, and with the substitutions $x \longmapsto x + 1, x \longmapsto x-1, x \longmapsto x+2, x \longmapsto x-2$ to try and show it is irreducible, but that didn't work out. Clearly, there's no linear factor of $x^6 + 108$ in $\mathbb{Q}$, since $x^6 + 108$ doesn't have a root in $\mathbb{Q}$. Showing there's no two cubic polynomials in $\mathbb{Q}[x]$ that multiply to $x^6 + 108$, or no cubic and quadratic polynomial in $\mathbb{Q}[x]$ that multiply to $x^6 + 108$ also quickly becomes a hassle. Is there an easier way to determine this more quickly?
Thanks!
Best Answer
Hint Since the polynomial $$x^6 + 108$$ is monic, if the field factors over $\Bbb Q$, it factors over $\Bbb Z$. Reducing modulo $7$, we have $$x^6 + 108 \pmod 7 \equiv x^6 - 4 \pmod 7 \equiv (x^3 + 2) (x^3 - 2) \pmod 7 .$$
Thus, if $x^6 + 108$ factors over $\Bbb Q$, it factors as a product of two cubics.