If $\alpha , \beta$ are the roots of $a{x}^2+bx+c=0$ then find the value of 

i) $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$     $ii) {\alpha }^4{\beta }^7+{\alpha }^7{\beta }^4$    $iii) \frac{{\alpha }^2+{\beta }^2}{{\alpha }^{-2}+{\beta }^{-2}}$   $iv) \frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}$

If $\alpha , \beta$ are the roots of the equation $a{x}^2+bx+c=0$

then $\alpha +\beta =\frac{-b}{a}, \alpha \beta =\frac{c}{a} i.e s_1=\frac{-b}{a}, s_2=\frac{c}{a}$

$i)$ $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$

$$\begin{gathered} \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}=\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}=\frac{s_1^2-2s_2}{s_2^2} \\ =\frac{{\left({\displaystyle \frac{-b}{a}}\right)}^2-2\left({\displaystyle \frac{c}{a}}\right)}{{\left(\frac{c}{a}\right)}^2}=\frac{{\displaystyle \frac{b^2}{a^2}}-{\displaystyle \frac{2c}{a}}}{{\displaystyle \frac{c^2}{a^2}}}=\frac{b^2-2ac}{c^2} \end{gathered}$$

$ii) {\alpha }^4{\beta }^7+{\alpha }^7{\beta }^4$

$$\begin{gathered} {\alpha }^4{\beta }^7+{\alpha }^7{\beta }^4= {\alpha }^4{\beta }^4\left( {\alpha }^3+{\beta }^3\right) \\ ={\left(\alpha \beta \right)}^4\left[{\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right] \\ ={\left(\frac{c}{a}\right)}^4\left[{\left(\frac{-b}{a}\right)}^3-3\left(\frac{c}{a}\right)\left(\frac{-b}{a}\right)\right]=\frac{c^4}{a^4}\left[-\frac{b^3}{a^3}+\frac{3bc}{a^2}\right] \\ =\frac{c^4}{a^4}\left[\frac{-b^3+3abc}{a^3}\right]=\frac{b{c}^4\left[3ac-b^2\right]}{a^7} \end{gathered}$$

$iii) \frac{{\alpha }^2+{\beta }^2}{{\alpha }^{-2}+{\beta }^{-2}}$

$$\begin{gathered} \frac{{\alpha }^2+{\beta }^2}{{\alpha }^{-2}+{\beta }^{-2}}=\frac{{\alpha }^2+{\beta }^2}{{\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}}=\frac{{\alpha }^2+{\beta }^2}{{\displaystyle \left(\frac{{\displaystyle {\alpha }^2+{\beta }^2}}{{\displaystyle {\alpha }^2{\beta }^2}}\right)}} \\ ={\alpha }^2{\beta }^2={\left(\alpha \beta \right)}^2={\left(\frac{c}{a}\right)}^2=\frac{c^2}{a^2} \end{gathered}$$

$iv) \frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2}$

$$\begin{gathered} \frac{\alpha }{{\beta }^2}+\frac{\beta }{{\alpha }^2} \\ =\frac{{\alpha }^3+{\beta }^3}{{\alpha }^2{\beta }^2}=\frac{1}{{\left(\alpha \beta \right)}^2}\left[{\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right] \end{gathered}$$

$$\begin{gathered} =\frac{1}{{\left({\displaystyle \frac{c}{a}}\right)}^2}\left[{\left(\frac{-b}{3}\right)}^3-3\frac{c}{a}\left(\frac{-b}{a}\right)\right] \\ =\frac{a^2}{c^2}\left[\frac{-b^3}{a^3}+\frac{3bc}{a^2}\right]=\frac{a^2}{c^2}\left[\frac{-b^3+3abc}{a^3}\right] \\ =\frac{3abc-b^3}{a{c}^2} \end{gathered}$$