Eliminating parameters from polynomial relations Let $p =x^3+y^3+z^3$, $q= x^2y+y^2 z+z^2x$, $r=xy^2+yz^2+zx^2$, and $s=xyz$. I want to find some non-zero polynomial in $\phi$ in $p,q,r,s$ such that $\phi(p,q,r,s)=0$; that is, to eliminate $x,y,z$. I have $p^2-2qr-2ps+6s^2=x^6+y^6+z^6$ and $p^3-3q^3-3r^3-24s^3+18qrs=x^9+y^9+z^9$, but at this point it seems like I'm going in circles. The PDF I'm reading is on ring actions, if that gives any context.

Given the polynomials
$$ p \!=\! x^3\!+\!y^3\!+\!z^3,\;\; q\!=\! x^2y\!+\!y^2 z\!+\!z^2x,\;\;
 r\!=\!xy^2\!+\!yz^2\!+\!zx^2,\;\; s\!=\!xyz \tag{1} $$
then we have the syzygy
$$ 0 = q^3 + r^3  + 9 s^3 + p^2 s + 3 p s^2 - p q r- 6 r q s . \tag{2} $$
I used a general purpose tool for finding algebraic relations to find it.
In this particular case, there is a homogeneous cubic polynomial
with $20$ different possible monomials. If needed, you can use the
undetermined coefficients method and solve for the integer coefficients.
My tool just automates the method using a PARI/GP function I wrote.
