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Consider $G:=\operatorname{GL}_3(\Bbb C)$ acting from the right on $R=\Bbb C[x_0,x_1,x_2]_3$ by linear substitution, i.e. we let $x=(x_0,x_1,x_2)^T$ and the action of $A\in G$ on $f\in R$ yields the polynomial $f(Ax)$. For polynomials $f,g\in R$ we write $f\sim g$ if there is a matrix $A\in G$ such that $f(Ax)=g(x)$. It is known that for every $f\in R$, there is an $a\in\Bbb C$ such that $f \sim f_a$, where $$ f_a := x_0^3 + x_1^3 + x_2^3 + a\cdot x_0x_1x_2. $$ Unfortunately for the non-German speakers, my only reference for this is the book Geometrische Methoden in der Invariantentheorie by Hanspeter Kraft, but I am reasonably sure this can be found elsewhere.

What I can not find is a proof for the following statement:

There are only finitely many pairs $(a,b)\in\Bbb C^2$ with $a\ne b$ and $f_a\sim f_b$.

For example, if there is some third root of unity $\zeta\in\Bbb C$ with $b=\zeta a$, then $f_b=f_a(\zeta x_0, x_1, x_2)$ but not much more can happen.

I am quite sure the statement is true, but I would be very interested to see a (direct/elementary) proof because I would like to prove something very similar and I am stuck.

Amendment. I am aware that one way to prove this is by showing that some polynomial in $a$ is a $G$-invariant function on $R$. I am looking for something even more elementary, because finding an invariant in my case seems hard. I am aware that the question is a bit fuzzy, and I deleted it because of that flaw - I undeleted it now because a fellow user asked me to, in private communication.

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    This beautiful family of cubics is called the _Hesse pencil_. For your question, it suffices to show that there are only finitely many such pairs for which the $j$-invariants of $f_a$ and $f_b$ coincide. A formula for the $j$-invariant of $f_a$ is (more or less) given on p.5 of the nice paper https://arxiv.org/pdf/math/0611590v3.pdf by Artebani--Dolgachev.2017-01-17
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    @Nefertiti: Thanks, this is of course the right way to do it, but it's not helping me. My problem is not *actually* with ternary cubics, I am dealing with much more complicated polynomials. I was hoping to gain some perspective by looking at an elementary proof for the classification of cubics. However, the j-invariant is not it, because it is an invariant that is very specific to ternary $3$-forms.2017-01-26
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    Hey @JeskoHüttenhain, sorry for taking so long to reply, but I indeed have been thinking about your question still! I'll post my thoughts sometime in the next few days when I get the time.2017-02-09

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