Let $E$ be the set of cubic ternary forms $$E=\{ P=aX^3+bY^3+cZ^3+3dX^2Y+3eY^2Z+3fZ^2X+3gXY^2+3hYZ^2+3iZX^2+6jXYZ : (a,b,c,d,e,f,g,h,i,j) \in \mathbb{C}^{10} \}.$$
Consider the subset $F$ of forms in $E$ which are the product of three linear forms. By definition, $F$ is the image of $$\phi:(m_{ij})_{1\leq i,j\leq 3} \mapsto (m_{11}X+m_{12}Y+m_{13}Z)(m_{21}X+m_{22}Y+m_{23}Z)(m_{31}X+m_{32}Y+m_{33}Z).$$
By a Theorem of Chevalley, $F$ is a constructible set. Let $\bar{F}$ be its Zariski closure.
Question : Can we find explicit polynomials defining $\bar{F}$, and how ?
Unfortunately (for me), I know little of algebraic geometry. Here some remarks:
1) An open subset of the image is a homogeneous space $GL(3,\mathbb{C})/Stab(XYZ)$, the stabilizer being (up to finite index) the set of diagonal matrices of determinant one. Thus the expected dimension of $\bar{F}$ is 9-2=7.
2) I am aware that cubic ternary forms have two invariants $S,T$, called the Aronhold invariants, homogeneous of degree $4$ and $6$, that generates the algebra of invariant polynomial under the action of SL(3) by substitution. (For a generic element $P$ of $E$, these two polynomials seem to be related to the $g_2$ and $g_3$ coefficients of the underlying elliptic curve described by $P(X,Y,1)=0$, although the exact relationship isn't clear to me). Namely, using the coordinates above $$S = a*g*e*c - a*g*h^2 - a*j*b*c + a*j*e*h + a*f*b*h - a*f*e^2 - d^2*e*c + d^2*h^2 + d*i*b*c - d*i*e*h + d*g*j*c - d*g*f*h - 2*d*j^2*h + 3*d*j*f*e - d*f^2*b - i^2*b*h + i^2*e^2 - i*g^2*c + 3*i*g*j*h - i*g*f*e - 2*i*j^2*e + i*j*f*b + g^2*f^2 - 2*g*j^2*f + j^4.$$ and an awfully long formula for $T$. I checked (or rather, I asked Sagemath to check) that we have on $F$ the relationship
$$64*S^3=T^2.$$ I guessed this because of the discriminant of elliptic curves : the ones in $F$ are fairly degenerate (union of three lines), so have discriminant zero. That's one equation, but i miss (a least) two others to describe a dimension 7 algebraic variety. Moreover, it is of degree 12, so I suspect it is consequence of smaller degree relationships.
3) The question here seems related :
This seems worth an answer: by Aronhold(1849). Given a form, the Hessian refers to the determinant of the matrix of second partial derivatives. Given a ternary cubic form $f,$ the Hessian $ \mathcal H$ is also a ternary cubic form.
Theorem. $f$ is completely reducible if and only if there is a scalar $\lambda \in \mathbb C$ such that $$ \mathcal H = \lambda f \; \; . $$ Here $\lambda$ turns out to just be double the determinant of the 3 by 3 matrix of linear coefficients of the three linear factors.
Source Gary Brookfield pdf
There is an 1852 book by Salmon that seems to discuss this or generalizations, I have requested a library copy (of one of the reprint editions)
Look at that, there is a 2000 book by A. Schinzel called Polynomials with Special Regard to Reducibility. I requested that also, should be easier to read. Apparently, on page 213, there is a discussion of when the form is not completely reducible but is reducible, which would be a linear times an irreducible quadratic form.
For example, the form I did at How to show that if $x, y, z$ are rational numbers satisfying $(x + y + z)^3 = 9(x^2y + y^2z +z^2x)$, then $x = y = z$ has Hessian determinant $0.$