$P(\sqrt[3]{x})$ won't work here

First off, we notice by Vieta's that $pqr=-\dfrac{1}{2}$ . Thus, $pq=-\dfrac{1}{2r}$ , $qr=-\dfrac{1}{2p}$ , and $rp=-\dfrac{1}{2q}$ .

Plugging these values in, we get $\left(-\dfrac{1}{2r}\right)^3+\left(-\dfrac{1}{2p}\right)^3+\left(-\dfrac{1}{2q}\right)^3$ or $-\dfrac{1}{8}\left(\dfrac{1}{p^3}+\dfrac{1}{q^3}+\dfrac{1}{r^3}\right)$

How do we find the value of $\dfrac{1}{p^3}+\dfrac{1}{q^3}+\dfrac{1}{r^3}$ though? If we could find another polynomial with roots $\dfrac{1}{p},\dfrac{1}{q},\dfrac{1}{r}$ then we could find it. Thankfully, this polynomial is simply $Q(x)=x^3P\left(\dfrac{1}{x}\right)=x^3+5x^2+3x+2$

Thus all we need to do is find the sum of the cubes of the roots of $Q(x)$ . This sounds like a job for Newton Sums. Let $P_n=\alpha^n+\beta^n+\gamma^n$ , where $\alpha,\beta,\gamma$ are the roots of $Q(x)$ .

We have that: $P_1+5=0$ $P_2+5P_1+6=0$ $P_3+5P_2+3P_1+6=0$

Solving for $P_3$ , we get that $P_3=-86$ . Thus, $\alpha^3+\beta^3+\gamma^3=-86$ , so $\dfrac{1}{p^3}+\dfrac{1}{q^3}+\dfrac{1}{r^3}=-86$ , and $p^3q^3+q^3r^3+r^3p^3=-\dfrac{1}{8}\left(\dfrac{1}{p^3}+\dfrac{1}{q^3}+\dfrac{1}{r^3}\right)=\dfrac{43}{4}$ thus our final answer is $\boxed{47}$ .

By Vieta's we have: $p+q+r=-\dfrac{3}{2}, ~pq+qr+rp=\dfrac{5}{2}, ~ pqr=-\dfrac{1}{2}$ . Now note that:

$\begin{aligned} (pq)^2+(qr)^2+(rp)^2&=& (pq+qr+rp)^2-2(pq\cdot qr+qr\cdot rp + rp\cdot pq) \\ &=& (pq+qr+rp)^2 - 2pqr(p+q+r) \\ &=& \left(\dfrac{5}{2}\right)^2 - 2\cdot \left(-\dfrac{1}{2}\right)\cdot \left(-\dfrac 3 2\right) \\ &=& \dfrac{19}{4} \end{aligned}$

So we have:

$\begin{aligned} (pq)^3+(qr)^3+(rp)^3 &=& \left(pq+qr+rp\right)\left\{(pq)^2+(qr)^2+(rp)^2-pq\cdot qr-qr\cdot rp-rp\cdot pq\right\} +3p^2q^2r^2 \\ &=& \left(pq+qr+rp\right)\left\{(pq)^2+(qr)^2+(rp)^2-pqr(p+q+r)\right\}+3(pqr)^2 \\ &=&\dfrac 5 2 \left\{\dfrac{19}{4}-\left(-\dfrac 1 2\right)\left(-\dfrac 3 2\right)\right\}+3\left(-\dfrac 1 2\right)^2 \\ &=& \boxed{\dfrac{43}{4}} \end{aligned}$

Hence the answer is: $43+4=\fbox{47}$ .

From Vieta's formulas, $p+q+r=\frac{-3}{2}$ , $pq+pr+qr=\frac{5}{2}$ , and $pqr=\frac{-1}{2}$ .

We can rewrite the original expression as the following:

$p^3q^3+q^3r^3+r^3p^3$

$=(pq+pr+qr)^3-(3p^3q^2r+3p^3r^2q+3q^3p^2r+3q^3r^2p+3r^3p^2q+3r^3q^2p+6p^2q^2r^2)$

$=(pq+pr+qr)^3-6(pqr)^2-3pqr(p^2q+p^2r+q^2p+q^2r+r^2p+r^2q)$

$=(pq+pr+qr)^3-6(pqr)^2-3pqr[(p+q+r)(pq+pr+qr)-3pqr]$

We can now plug in our values for $p+q+r$ , $pq+pr+qr$ , and $pqr$ to get $\frac{43}{4}$ , so our answer is $\fbox{47}$ .