Given integers $A$, $B$, $C$, I would like to solve the following system of quadratic equations for integers $a$, $b$, $c$, $d$: $$ \begin{aligned} A &= b^2 - ac, \\ B &= ad - bc, \\ C &= c^2 - bd. \\ \end{aligned} $$
The equations are relating a binary cubic form $ax^3 + 3bx^2y + 3cxy^2 + dy^3$, to one of its 'covariants' (web1, wikipedia), a binary quadratic form $A x^2 + B xy + C y^2$. As you can see, it is easy to start from the cubic form and get the binary form, but it is not clear to me how to do the opposite.
Playing with this, I found that there are some nice 'linear' equations hidden in there $$ \begin{aligned} c A + b B + a C &= c (b^2 - ac) + b (ad - bc) + a (c^2 - bd) = 0, \\ d A + c B + b C &= d (b^2 - ac) + c (ad - bc) + b (c^2 - bd) = 0. \\ \end{aligned} $$
So I'm hoping there is some "nice" way to solve this with linear tools (extended Euclidean algorithm).
However there are also some rearrangements that isolate two of the variables in a quadratic equation $$ \begin{aligned} b^2 A + ab B + a^2 C &= b^2 (b^2 - ac) + ab (ad - bc) + a^2 (c^2 - bd) \\ &= b^2 (b^2 - ac) - ab(bc) + a^2 (c^2) \\ &= (b^2 - ac)^2 \\ &= A^2 \\ \end{aligned} $$
One infinite family of these binary quadratic forms is $$\langle U, V, U^2 \rangle$$ which arises from $$ a = 1, b = 0, c = -U, d = V $$
We see that this has order (dividing) three in the class group because Dirichlet's method of composition shows $$\langle U, V, U^2 \rangle^3 \equiv \langle U^3, V, 1 \rangle$$ which is the identity form.
One must adjust the middle coefficient to have this work out;
This is in chapter XII, pages 253-258 in Volume 3 of Dickson, History of the Theory of Numbers. He says Eisenstein described the invariants and covariants in 1844.
Apparently Pepin (1883) showed surjectivity for this when the discriminant of the quadratic form is $-4p$ and $p \equiv 3 \pmod 4$ is prime. I can't tell (yet) whether Eisensteins's map is surjective in general.