If $p+q+r=6,pq+pr+qr=8,pqr=2$, what is the value of: $$\frac{1}{(pq)^2-10r^2+300}+\frac{1}{(qr)^2-10p^2+300}+\frac{1}{(pr)^2-10q^2+300}$$
I tried to change $\frac{1}{(pq)^2-10r^2+300}$ to $\frac{1}{\frac{4}{r^2}-10r^2+300}$ but I don't know how to solve it.
Please provide a hint.
$p$, $q$ and $r$ are roots of the equation: $$x^3-6x^2+8x-2=0,$$ which by your work gives: $$\sum_{cyc}\frac{1}{p^2q^2-10r^2+300}=\sum_{cyc}\frac{r^2}{4-10r^4+300r^2}=$$ $$=\sum_{cyc}\frac{r^2}{2(r^3-6r^2+8r)-10r^4+300r^2}=\sum_{cyc}\frac{r}{16+288r+2r^2-10r^3}=$$ $$=\sum_{cyc}\frac{r}{8(r^3-6r^2+8r)+288r+2r^2-10r^3}=-\frac{1}{2}\sum_{cyc}\frac{1}{r^2+23r-176}.$$ Can you end it now?
For example, I got $$\prod_{cyc}(r^2+23r-176)=(pqr)^2+pqr(23(pq+pr+qr)+881(p+q+r)+24311)+$$ $$+30976(p+q+r)^2-176(pq+pr+qr)^2+712448(p+q+r)-$$ $$-4048(p+q+r)(pq+pr+qr)-155056(pq+pr+qr)-5451776=$$ $$=-1448162.$$