Determining the Galois group of a cubic without using discriminant

Perhaps not quite basic Galois theory, but definitely a standard part of basic algebraic number theory:

Assuming we've rearranged so that the (irreducible) cubic $f(x)$ is monic and has integer coefficients, if we can find a prime $p$ such that the reduction of $f$ mod $p$ factors as linear polynomial and irreducible quadratic, then there is a $2$-cycle in the global Galois group. The irreducibility implies presence of a $3$-cycle, so there's no choice left but that the Galois group is $S_3$.

Similarly, if the cubic has exactly one real root (and two complex), we also conclude that there's a $2$-cycle in the Galois group.

The same line of thinking applies to irreducible prime degree $q$ polynomials: the irreducibility implies existence of a $q$-cycle in the Galois group. If not all roots are real, and/or if there is some prime modulo which there is an irreducible quadratic factor, then the Galois group has to be $S_q$, since any $q$-cycle and a $2$-cycle generate the whole thing.