Take any cubic real polynomial $f$ with roots $p,q,r$.
Let $x = p^2 q + q^2 r + r^2 p$.
Let $y = p q^2 + q r^2 + r p^2$.
Let $z = p + q + r$.
Then $x+y$ and $xy$ are both symmetric in $p,q,r$ and hence can be expressed in terms of the coefficients of $f$. This implies that we can solve for $x,y$ via the appropriate quadratic equation.
But after that, is it possible to algebraically recover the roots $p,q,r$ from $x,y,z$? I think that the three equations uniquely determine them, but I do not see an easy or motivated way to continue. Is this approach viable or is it doomed to failure, and why?
This question was inspired by this post, whose author suggested such an approach but we could not see how to complete it to obtain the roots.
Suppose the cubic has the form
$$f(x) = t^3 + bt - c$$
shifted in advance so that the coefficient of $t^2$ is $0$.
Letting $p,q,r$ denote the roots, and let
\begin{align*} x &= p^2q + q^2r + r^2p\\[6pt] y &= pq^2 + qr^2 + rp^2\\[6pt] z &= p + q + r \end{align*}
Since the coefficient of $t^2$ is $0$, it follows that $z=0$.
Let $u = x + y$ and let $v = xy$.
Using Maple, I derived the following relations:
\begin{align*} &u = -3c\\[3pt] &v = b^3+9c^2\\[14pt] &x^2+(3c)x+(b^3+9c^2) = 0\\[3pt] &y^2+(3c)y+(b^3+9c^2) = 0\\[14pt] &c = -\frac{x+y}{3}\\[3pt] &b^3 = -(x^2 + xy + y^2)\\ \end{align*}
But recovering $p,q,r$ from $x,y$ is not really viable$\,-\,$you get a $9$-th degree equation (although it's effectively cubic):
$$27p^9+27(x+y)p^6-9(2x^2+xy+2y^2)p^3+(x+y)^3 = 0$$
and the same equation for the other two roots.