$\begin{array}{r}a x-a p \\ x ^ { 2 } + p x + 1 \longdiv { a x ^ { 3 } + b x + c } \\ \mathrm{ax}^3+\mathrm{apx}^2+a x \\ -\quad-\quad- \\ -\mathrm{apx}^2+(\mathrm{b}-\mathrm{a}) \mathrm{x}+\mathrm{c} \\ -\mathrm{apx}^2-\mathrm{ap}^2 \mathrm{x}-\mathrm{ap} \\ +\quad+\quad+ \\ \left(\mathrm{ba}+\mathrm{ap}^2\right) \mathrm{x}+\mathrm{ap}+\mathrm{c}\end{array}$
Since $x^2+P x+1$ is factor of $a x^3+b x+c$
$\therefore a p+c=0 \Rightarrow p=-\frac{c}{a}$ ...(i)
and $n-a+a p^2=0$
$\Rightarrow \mathrm{n}-\mathrm{a}+\mathrm{a} \frac{\mathrm{c}^2}{\mathrm{a}^2}=0 \quad[$ from (i) $]$
$\begin{aligned} & \Rightarrow a b-a^2+c^2=0 \\ & \Rightarrow a b=a^2-c^2\end{aligned}$