Maybe the roots are rational? I'll try that

Since $p$ , $q$ and $r$ are the roots of $f(x) = x^3+3x+8$ .
Therefore, $p+q+r=0$ , $pq+qr+rp=3$ and $pqr = -8$ .

Since $g(x) = \left( x - \frac {p^2+q^2}{r^2} \right) \left( x - \frac {q^2+r^2}{p^2} \right) \left( x - \frac {r^2+p^2}{q^2} \right)$

$\Rightarrow g(-1) = \left( -1 - \frac {p^2+q^2}{r^2} \right) \left( -1 - \frac {q^2+r^2}{p^2} \right) \left( -1 - \frac {r^2+p^2}{q^2} \right)$

We note that: $-1 - \dfrac {p^2+q^2}{r^2} = - \dfrac {r^2 + p^2+q^2}{r^2} = - \dfrac {(p+q+r)^2-2(pq+qr+rp)}{r^2}$

$= - \dfrac {0 - 2(3)}{r^2} = \dfrac {6}{r^2}$

$\Rightarrow g(-1) = \left( \dfrac {6}{r^2} \right) \left(\dfrac {6}{p^2} \right) \left(\dfrac {6}{q^2} \right) = \dfrac {6^3} {(pqr)^2} = \dfrac {216} {64} = \dfrac {27} {8} = \boxed {35}$

Lemma: The roots of a equation $x^3+px^2+qx+r=0$ are $a,b,c$ .

Then cubic equation whose roots are $\frac{a+b}{c}$ , $\frac{b+c}{a}$ and $\frac{c+a}{b}$ is $r(x+1)^{3}-pq(x+1)^{2}+p^{3}(x+1)-p^{3}=0$

Proof Let $y=\frac{b+c}{a}=\frac{-p-a}{a}$ (since $a+b+c=-p)$

Hence, $y=-\frac{p}{a}-1$ . So, $\frac{1}{a}=\frac{y+1}{-p}$

Since, $a$ is a root of the given equation,

$a^3+pa^2+qa+r=0$

Dividing both sides by $a^{3}$ ,we get

$1+p\frac{1}{a}+q\frac{1}{a^2}+r\frac{1}{a^3}=0$

Putting $\frac{1}{a}=\frac{y+1}{-p}$ and multiplying both sides by $-p^{3}$ , we get the required equation.

Solution of the Main Problem

First, we will find the equation whose roots are $p^2,q^2,r^2$

$p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp)=0-2\times3=-6$

$p^{2}q^{2}+q^{2}r^{2}+r^{2}p^{2}=(pq+qr+rp)^2-2pqr(p+q+r)=3^2=9$

$p^{2}q^{2}r^{2}=(-8)^2=64$

So, the equation whose roots are $p^2,q^2,r^2$ is $x^3+6x^2+9x-64=0$ ....(i)

From (i), we get the equation whose roots are $\frac{p^2+q^2}{r^2}$ , $\frac{q^2+r^2}{p^2}$ and $\frac{r^2+p^2}{q^2}$ (Using the lemma)

The equation is $-64(x+1)^3-54(x+1)^2+216(x+1)-216=0$

Hence, the required monic cubic is

$g(x)=(x+1)^3+\frac{54}{64}(x+1)^2-\frac{216}{64}(x+1)+\frac{216}{64}$

Hence $g(-1)=\frac{216}{64}=\frac{27}{8}$

hence, $a+b=27+8=35$