If $\alpha, \beta, \gamma$ are the roots of $x^3+2 x^2-3 x-1=0$ then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$



On squaring equation (ii), we get
$\begin{aligned}
& \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2+2 \alpha \beta \gamma(\alpha+\beta+\gamma)=9 \\
& \Rightarrow \alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2=9-2(1)(-2)=13
\end{aligned}$
Now
$\begin{aligned}
& \alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{\beta^2 \gamma^2+\gamma^2 \alpha^2+\alpha^2 \beta^2}{(\alpha \beta \gamma)^2} \\
& \alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{13}{1}=13
\end{aligned}$