Is there a way to understand whether there exist linear transformation that brings one cubic form of n variables to another form? In particular one of the examples I am interested in are two cubic forms in 4 variables:

$F_1=x_1 x_2 x_3+x_1 x_3 x_4+ x_1 x_2 x_4+ x_2 x_3 x_4$

$F_2=x_1 x_2 x_3 + x_1 x_3 x_4 + 2x_1 x_2 x_4$

If there is a way to understand whether such transformation exists is there an algorithm allowing to find this transformation?

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    $\begingroup$ surely, there is an algorithm reducing this question to solving a system of polynomial equations with unknowns being the entries of a 3x3 matrix. $\endgroup$ – Dima Pasechnik Sep 4 '19 at 9:06
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    $\begingroup$ You need to understand the orbits of the action of $\mathrm{GL}(4,K)$ on $S^3(K^4)$. For such questions you need to be more specific about the ground field. $\endgroup$ – YCor Sep 4 '19 at 9:07
  • $\begingroup$ In your example the answer is no. The surface $F_1=0$ has isolated singularities, while $F_2=0$ is reducible. $\endgroup$ – abx Sep 4 '19 at 9:21
  • $\begingroup$ @DimaPasechnik: To be fully precise, you would have to also encode the nonsingularity of the transformation matrix as a polynomial inequation. While there are algorithms for this using Grobner bases (you have to check whether a polynomial vanishes all over the zero set of an ideal), the question whether there are better algorithms than that is reasonable. Still, probably math.stackexchange material. $\endgroup$ – darij grinberg Sep 4 '19 at 15:25
  • $\begingroup$ @abx: Thank you for the suggestion. I was looking for some criteria like this. However could you please elaborate a bit. I would think that $F_2=0$ has node singularities at $(1,0,0,0)$ and its permutations just like $F_1=0$ has. Why is this not a case? $\endgroup$ – Anton Nedelin Sep 4 '19 at 16:05