If $x_1, x_2$ are the two roots of the quadratic equation $a{x}^2+bx+c=0$ Find the value of ${\left(a{x}_1+b\right)}^{-2}+{\left(a{x}_2+b\right)}^{-2}$ in terms of a, b, c

$x_1, x_2$ are roots of 

$a{x}^2+bx+c=0\Rightarrow x_1+x_2=\frac{-b}{a}$

$$\begin{gathered} \therefore x^2-x\left[\left({\alpha }^2+{\beta }^2\right)+\left({\alpha }^{-2}+{\beta }^{-2}\right)\right]+\left({\alpha }^2+{\beta }^2\right)\left({\alpha }^{-2}+{\beta }^{-2}\right)=0 \\ {x}_1{x}_2=\frac{c}{a} x_1 is root of \\ a{x}^2+bx+c=0 \Rightarrow a{x}_1^2+b{x}_1+c=0 \\ a{x}_{1_{ }}+b=\frac{-c}{x_1} \\ {\left(a{x}_1+b\right)}^{-2}={\left(\frac{-c}{x_1}\right)}^{-2}={\left(\frac{{\displaystyle -x_1}}{{\displaystyle c}}\right)}^{-2}=\frac{x_1^2}{c^2} \\ similarly {\left(a{x}_2+b\right)}^2=\frac{x_2^2}{c^2} \\ \therefore {\left(a{x}_1+b\right)}^{-2}+{\left(a{x}_2+b\right)}^{-2}=\frac{{\displaystyle x_1^2}}{{\displaystyle c^2}}+\frac{{\displaystyle x_2^2}}{{\displaystyle c^2}} \\ =\frac{{\displaystyle {\left(x_1+x_2\right)}^2}{\displaystyle -}{\displaystyle 2}{\displaystyle {{\displaystyle x}}_1}{\displaystyle {{\displaystyle x}}_2}}{c^2}=\frac{b^2-2ac}{a^2{c}^2} \end{gathered}$$