There are at least two ways to construct cyclic cubic extensions of $\mathbb{Q}$ as explained below. (A third one is given in the answer to an earlier question).
Given $A, B, C$ integers with $A\neq 0$ we put $$ \lambda=-\frac{C^2+27B^2}{4A^3} $$ Then, if $(a,b,c)=\lambda(A,B,C)$, we see that $c^2=-4a^3-27b^2$. Hence, the discriminant of $T^3+aT+b$ is a square in $\mathbb{Q}$. So the simple extension $\mathbb{Q}[z]$ obtained by adjoining a root $z$ of this polynomial is normal and cyclic. (Assume that $A$ and $B$ are chosen so that this polynomial is irreducible which is not difficult.)
Given a positive integer $k$ such that $3$ divides the order of $(\mathbb{Z}/(k))^{*}$, there is a subgroup $H$ of this group that has index $3$. Choosing one such $H$, we see that if $E=\mathbb{Q}[\zeta_k]$, then $F=E^{H}$ is a cyclic cubic extension of $\mathbb{Q}$.
According to the Kronecker-Weber Theorem, every construction in (1) is associated with some construction in (2).
Question: Given $a$ and $b$, is there an algorithm to determine $k$? (The link referenced above can perhaps be used to determine an $a$ and $b$ given $H$.)
To begin with even a "brute force" algorithm based on giving a bound on $k$ in terms of $a$ and $b$ would be nice.
A specific case (which motivated this question) is when $(a,b)=(-7,-7)$. Then, serendipity led one to the solution $k=7$. This pair $(a,b)$ is different from the one produced by the referenced link. (Of course, multiple $(a,b)$ pairs could give the same extension.)
