When do the splitting fields of two cubic polynomials coincide?

$\newcommand{\Kbar}{{\overline K}} $Let $K$ be a field with a fixed algebraic closure $\Kbar$. Let $L/K$ be a Galois extension of degree $6$ in $\Kbar$ with Galois group $S_3$ (the symmetric group on three symbols). By this answer the field $L$ is the splitting field in $\Kbar$ of an irreducible cubic polynomial $$ f(X)=X^3+p X+q $$ with $p,q\in K$.

Now let $$ f'(X)=X^3+p' X+ q' $$ be another cubic polynomial.

Question 1. Under what conditions on $p,q,p',q'$ do the splitting fields in $\Kbar$ of the irreducible cubic polynomials with Galois group $S_3$ $$ f(X)=X^3+p X+q \quad\ \text{and}\quad\ f'(X)=X^3+p'X+q'$$ coincide?

Question 2. The same for the irreducible cubic polynomials $$ f(X)=X^3+p X+q \quad\ \text{and}\quad\ f'(X)=X^3+p'X+q'$$ with Galois group $C_3$ (a cyclic group of order $3$).

If there is no satisfactory answer for a general field $K$, then I will be happy to have answers in the case where $K$ is a global or local field.

These questions were asked in MSE for $K={\mathbb Q}$. However, in my opinion the answers there are not quite satisfactory, so I ask the questions here in MathOverflow.

Here is a sketch of a somewhat brutal algorithm based on Frobenius density as suggested in Peter Mueller's comment, for the case where $K$ is cyclic cubic over $\mathbb{Q}$ (i.e. $\mathrm{Gal}(K/\mathbb{Q}) = C_3$.

First, for two extensions $K,K'$ to agree, it is necessary that their discriminants agree. I found this nice paper which determines the discriminant of a cubic field purely in terms of a presenting polynomial, so this can be done.

Next, as in Peter Mueller's comment, $K$ and $K'$ agree if the splitting behavior of all rational primes not dividing the discriminant is the same. For any finite number of primes you can determine the splitting behavior using the paper linked above, but to extend to infinitely many we need another ingredient, namely Artin reciprocity.

Artin reciprocity tells you (among other things) that the splitting behavior in $K$ of a prime $\ell$ not dividing the discriminant $D$ only depends on the residue class of $\ell$ mod $D$. (This is where we use that $K/\mathbb{Q}$ is cyclic!) So it suffices to compare the splitting behavior in $K$ and $K'$ of one representative $\ell$ of every residue class in $(\mathbb{Z}/D)^\times$. This last step can probably be sped up considerably, since what we're really computing here is a homomorphism $(\mathbb{Z}/D)^\times \to C_3$, so for example the residue classes which split completely are closed under multiplication.

Another approach to this question is via Galois theory (and in particular resolvent polynomials). The following procedure works in characteristic zero to answer both questions with a finite computation. Assume that $f(X)$ and $f'(X)$ are irreducible in $K[X]$.

The Galois group of $f(X) f'(X)$ is isomorphic to a subgroup of $S_{3} \times S_{3}$, and there is a unique index $6$ subgroup of this group with the property that the Galois group of $f(X) f'(X)$ is contained in this subgroup if and only if the splitting fields of $f(X)$ and $f'(X)$ are the same. It is possible to compute a degree $6$ resolvent for this subgroup: it is $$ g(X) = X^{6} - 6pp' X^{4} - 27qq' X^{3} + 9p^{2} (p')^{2} X^{2} + 81 pp'qq' X - 4p^{3} (p')^{3} - 27 p^{3} (q')^{2} - 27 q^{2} (p')^{3}. $$ (This may be well known, but my co-authors and I worked this out on page 19 of this paper, with the help of Magma).

If $f(X)$ and $f'(X)$ have the same splitting field over $K$, then $g(X)$ will have a root in $K$. Conversely, if $g(X)$ has a root in $K$ and the discriminant of $g(X)$ is nonzero, then the splitting fields of $f(X)$ and $f'(X)$ are the same.

The discriminant of $g$ is zero if and only if one of the following occurs:

Some remarks which are too long for a comment:

Note that you can assume this form of the cubic only if the characteristic of $K$ isn't $3$. In addition, if $p\ne0$, you may assume $q=p$ (by passing to $p^3f(qX/p)/q^3$). This might help to formulate (potential) answers in a simpler fashion. Also, if $p=q$, then $f$ will have Galois group $C_3$ if and only if $-4p-27$ is a square.

One primitive criterion of course is: $f$ and $f'$ will have the same splitting field if and only if $f$ has a root in $K[X]/(f')$ and $f'$ has a root in $K[X]/(f)$.

If $K=\mathbb Q$, then by Frobenius density theorem $f$ and $f'$ have the same splitting field if and only if $f$ and $f'$ have the same factorization pattern modulo all primes which do not divide the discriminant of $f$ or $f'$.