How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $ 
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me in finding the value.
Any good hint is welcomed.

Alternative hint:  derive the equation whose roots are $y_1=(a\alpha + b)^{-2}$ and $y_2 = (a\beta + b)^{-2}\,$.
Let $y=(ax+b)^{-2}\,$, then $\;\displaystyle ax+b=\pm \frac{1}{\sqrt{y}} \;\;\iff\;\; x = \frac{-b \pm \frac{1}{\sqrt{y}}}{a} = \frac{-b\sqrt{y}\pm1}{a\sqrt{y}}\,$.
Substituting back into the original equation:
$$\require{cancel}
\begin{align}
a\,\frac{b^2 y \mp 2b\sqrt{y}+1}{a^2 y} + b\,\frac{-b \sqrt{y} \pm 1}{a \sqrt{y}}+c = 0 \;\;&\iff\;\; \cancel{b^2 y} \mp 2b\sqrt{y}+1 -\cancel{b^2 y} \pm b\sqrt{y}+acy = 0 \\
 &\iff\;\; acy \mp b\sqrt{y}+1=0 \\
 &\iff\;\; (acy+1)^2 = b^2y \\
 &\iff\;\; a^2c^2y^2 +(2ac-b^2)y+1=0
\end{align}
$$
Then $\displaystyle(a\alpha + b)^{-2} + (a\beta + b)^{-2} =y_1+y_2=\frac{b^2-2ac}{a^2c^2}$ by Vieta's relations.
