I'm wondering how to show that $f = X^6 - 6125 \in\mathbb{Q}[X]$ is irreducible. Since $f$ is monic I tried to look at it in $\mathbb{Z}[X]$ and even further look at it as
$$ \bar f =X ^6+1 \in \mathbb{F}_3[X] $$
$\bar f$ has no roots in $\mathbb{F}_3$, but it may be written as a product of polynomials $g, h$ of degree (2, 4) or (3, 3). I'd prefer not to determine the irreducible polynomials in $\mathbb{F}_3[X]$ if degree 2, 3, 4. Is there an easier way to approach this problem?
On
It is irreducible modulo $p = 13$, where it reads $f = x^{6} + 2$.
Clearly this polynomial has no root in $F = \operatorname{GF}(13)$, as the only 6-th powers in $F^{*}$ are $1, -1$.
Then it is only a matter to check that $x$ has period $6 \cdot 12 = 72$ modulo $f$, as $-2$ has period $12$ in $F^{*}$.
If $f$ could be factored as a product of polynomials of lesser degrees, the possibilities are $6 = 2 + 2 + 2 = 3 + 3 = 2 + 4$.
But $72$ does not divide $p^{2} -1$, $p^{3} - 1$, $\operatorname{lcm}(p^{2} - 1, p^{4} - 1)$.
We may check that such polynomial is irreducible over $\mathbb{F}_{13}$, hence it is irreducible over $\mathbb{Q}$, too.
Indeed, over $\mathbb{F}_{13}$ the polynomial $x^6+1$ factors as $\prod_{k=1}^{6}(x-q_k)$ where the set of $q_k$s is the set of non-quadratic residues $\!\!\pmod{13}$, but $x^6-6125=x^6-2$ and $2$ is neither a quadratic or a cubic residue $\!\!\pmod{13}$, such that the degree of the splitting field of $p(x)$ over $\mathbb{F}_{13}$ is six.