Let $F(x,y) = a_3 x^3 + a_2x^2 y + a_1 xy^2 + a_0 y^3$ be a binary cubic form, say with real coefficients. Put $H(x,y) = H_F(x,y)$ for the *Hessian covariant* of $F$, defined by

$$\displaystyle H_F(x,y) = \frac{1}{4} \begin{vmatrix} F_{xx} & F_{xy} \\ F_{xy} & F_{yy} \end{vmatrix},$$

and put

$$\displaystyle G(x,y) = G_F(x,y) = \begin{vmatrix} F_x & F_y \\ H_x & H_y \end{vmatrix}.$$

$H,G$ are both covariants of $F$ under the substitution action of $\operatorname{GL}_2$, and in particular, $H, G, F$ and the invariant $\Delta(F)$, the discriminant of $F$, generate the ring of polynomial covariants. They are connected by a single syzygy, given by

$$\displaystyle 4H(x,y)^3 + G(x,y)^2 = -27 \Delta(F) F(x,y)^2.$$

As can be verified by immediate calculation, we have

$$\displaystyle \Delta(G) = 729 \Delta(F)^3,$$

which is a perfect cube.

My question is, suppose that $G$ is a binary cubic form with integer coefficients satisfying $\Delta(G) = 729 n^3$ for some non-zero integer $n$. Does it follow that $G$ is the $G_F$-covariant for some binary cubic form $F$ with integer (rational) coefficients? If not, what is a counter-example, and what would be a stricter condition that guarantees $G$ is such a covariant?

On a related note, if $G'$ is the non-trivial cubic covariant of $G_F$, then $G' = -729 \Delta(F)^2 F(x,y)$.