For $(\alpha,\beta)\in\mathbb{R}^2$ be fixed, consider $f(x)=x^3-\alpha x+\beta$. Show that for any $(x,y)\in\mathbb{R}^2$, $x\not=y$, there exists a unique $(\alpha,\beta)\in\mathbb{R}^2$ such that $f(x)=f(y)=0$, and the function $F:\mathbb{R}^2\to\mathbb{R}^2$ that maps $(x,y)$ to $(\alpha,\beta)$ maps at most six different values to a single value.
My thought was that $f(x)=f(y)=0$ is equivalent to the matrix equation $\begin{bmatrix} -x & 1\\ -y & 1\end{bmatrix}\begin{bmatrix}\alpha\\ \beta\end{bmatrix}=\begin{bmatrix}-x^3\\-y^3\end{bmatrix}$. Since $x\not=y$, the matrix is invertible with unique solution $\begin{bmatrix}\alpha\\\beta\end{bmatrix}=\begin{bmatrix}y^2+xy+x^2\\xy(x+y)\end{bmatrix}$.
Is this approach correct, and are there alternatives? Also, I don't see why the function $F$ maps at most six different values to one.
Hint (for the second part): for given $\alpha,\beta\,$, the equation $\,z^3 - \alpha z + \beta = 0\,$ has roots $\,u,v,-u-v\,$. There are (at most) $\,6\,$ ways to choose $\,x,y\,$ from among them, in other words there are (at most) $\,6\,$ pairs $\,(x,y)\,$ which map to the same $\,\alpha,\beta\,$. There may be less than $\,6\,$ if the roots are not all real, or not all distinct.