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.
 A: 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.
A: Multiply $r$ and $q$
\begin{eqnarray*}
rq &=& \sum x^3y^3+ 3(xyz)^2+xyz \sum x^3 \\
&=&\sum x^3y^3+ 3s^2 +sp. \\
\end{eqnarray*}
So
\begin{eqnarray*}
\sum x^3y^3= rq -3s^2 -sp \\
\end{eqnarray*}
Now multiply this by $p$
\begin{eqnarray*}
p(rq-sp-3s^2) = \sum_{perms} x^6y^3+ 3(xyz)^3 \\
 \sum_{perms} x^6y^3= p(rq-sp-3s^2) -3s^3 \\
\end{eqnarray*}
Now cube $q$
\begin{eqnarray*}
q^3 = \sum_{cyc} x^6 y^3 + 3xyz \sum x^3 y^3 + 3(xyz)^2 \sum x^3 + 6(xyz)^3 \\
= \sum_{cyc} x^6 y^3 + 3 s(rq-3s^2-sp) +3s^2p+6s^3.
\end{eqnarray*}
Do the same for $r$ add and then eliminate using $\sum_{cyc} x^6y^3 +\sum_{cyc} x^3y^6=\sum_{perms} x^6y^3$.
This gives 
\begin{eqnarray*}
q^3 - 3 srq +3s^3 +r^3 - 3 srq +3s^3 = pqr-sp^2 -3s^2p -3s^3 \\
\color{red}{q^3+r^3+9s^3 +sp^2+3s^2p-pqr-6srq=0}.
\end{eqnarray*}
