Consider $P(x,y)$ a homogenous polynomial of degree $3$ in two variables (a binary cubic). To it we associate first the $2\times 2$ matrix
$$\frac{1}{2}\operatorname{Hess}(P) = \frac{1}{2}\cdot\left( \begin{matrix} \frac{\partial^2 P}{\partial x^2} & \frac{\partial^2 P}{\partial x \partial y } \\ \frac{\partial^2 P}{\partial y \partial x } & \frac{\partial^2 P}{\partial y^2 } \end{matrix} \right)$$ with entries linear forms in $x$, $y$, and then
$$D = \det (\frac{1}{2}\operatorname{Hess}(P))= (x, y) \cdot M \cdot (x,y)^t$$ a quadratic form in $x$, $y$, where $M$ is a uniquely defined symmetric matrix (with constant entries). The matrix $$N \colon = \left( \begin{matrix} 0 & -1\\ 1 & 0 \end{matrix} \right)\cdot M $$
is a matrix of trace $0$ (since $M$ is symmetric). Let also $\Delta$ be the discriminant of the binary form $P(x,y)$ ( which equals the discriminant of the polynomial $P(x,1)$)
Show that the $2\times 2 $ matrix
$$T = -\frac{1}{2} I_2 + \frac{N}{\sqrt{\Delta}}$$
is a transformation of order $3$ invariating $P$, that is
$$P(T\cdot \left ( \begin{matrix} x \\ y \end{matrix} \right)) = P(\left ( \begin{matrix} x \\ y \end{matrix} \right))$$
Notes:
The discriminant $\Delta$ of $P$ can also be defined by the equality $N^2 = -\frac{3}{4} \Delta \cdot I_2$ so indeed $T$ is of order $3$ (with characteristic polynomial $\lambda^2 + \lambda + 1$)
Notice that $T^2 = - \frac{1}{2} I_2 - \frac{N}{\sqrt{\Delta}}$, so we can choose any determination of the square root $\sqrt{\Delta}$.
We get the expression of the roots of the cubic in terms of one root. That is, if $x$ a root, then $T \cdot x$, $T^2 \cdot x$ are the other two roots, where $T$ is seen as a homographic transformation
Related to this question
Probably a classical result, so provided as a reference.
Any feedback would be appreciated!
