An algebra problem by MAMUNUR RASHID

The remainder when $f(x)$ is divided by $x-3$ is $p^3-q^3$ , hence

$\begin{aligned} f(3) &= p^3-q^3 \\ 9(p-q) + 3(p^2-q^2) + (p^3-q^3) &= p^3-q^3 \\ p+q &= -3 \quad (p-q\neq 0) \end{aligned}$

By Vieta's Formula, $\alpha \beta = \frac{p^3-q^3}{p-q} = p^2+pq+q^2$ and $\alpha + \beta = -\frac{p^2-q^2}{p-q}=-(p+q)=3$ .

$\begin{aligned} \alpha \beta - (\alpha + \beta)^2 &= 5 \\ p^2+pq+q^2 - 3^2 &= 5 \\ (p+q)^2 - pq - 9 &= 5 \\ pq &= -5 \end{aligned}$

The answer is $(p+q)^{-pq} = (-3)^5=-243$ .