Question from pathfinder for Olympiad mathematics 2 If $p$, $q$, $r$ are the real roots of equation $x^3-6x^2+3x+1=0$, determine the possible value of
$p^2q+q^2r+r^2p$.
My Attempt:
$p+q+r=6     (1)$
$pq+qr+pr=3  (2)$
$pqr=-1      (3)$
Multiplying (1) and (2) and substituting (3) in it...
I came close but still unable to find solution...
Please Help

Since $v_1=p^2q+q^2r+r^2p$ is not a symmetric polynomial, there might be multiple values. Another possible value is $v_2=p^2r+q^2p+r^2q.$
Now $v_1+v_2$ and $v_1v_2$ are symmetric. So we can express those values in terms of $p+q+r,$ $pq+pr+qr,$ and $pqr.$
Then solve for the two values, $v_1,v_2.$
I’ll do the harder case. Writing $s_1=p+q+r,$ $s_2=pq+pr+qr,$ and $s_3=pqr,$ we have:
$$\begin{align}v_1v_2=&(p^3+q^3+r^3)s_3+3s_3^2\\&+\left((pq)^3+(qr)^3+(rs)^3\right)
\end{align}$$
And $$\begin{align}p^3+q^3+r^3=&s_1^3-3s_1s_2+3s_3\end{align}$$
So that leaves us the last term, $(pq)^3+(pr)^3+(qr)^3.$ One clever way is to write it as:
$$(pq)^3+(pr)^3+(qr)^3=s_3^3\left(\frac1{p^3}+\frac1{q^3}+\frac1{r^3}\right)$$
then reuse the formula for the sum of three cubes we used previously, with $s_1’=s_2/s_3,$ $s_2’=s_1/s_3,$ $s_3’=1/s_3.$
So we have:
$$(pq)^3+(pr)^3+(qr)^3=s_2^3-3s_1s_2s_3+3s_3^2$$
So:
$$\begin{align}v_1v_2&=\left(s_1^3-3s_1s_2+3s_3\right)s_3+3s_3^2+(s_2^3-3s_1s_2s_3+3s_3^2)\\
&=s_1^3s_3-6s_1s_2s_3+s_2^3+9s_3^2\end{align}$$
Since $s_1=6,s_2=3,s_3=-1,$ this means: $$v_1v_2=-6^3+6\cdot 6\cdot 3 +3^3+9=-72.$$
I also get $v_1+v_2=21.$ So:
$$v_i=\frac{21\pm \sqrt{729}}{2}=-3\text{ or }24.$$
A numerical check using the roots provided by Wolfram Alpha indicate this is probably right.
