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

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3+2x^2-3x-1=0\\ \alpha +\beta +\gamma =-2,\alpha \beta +\beta \gamma +\gamma \alpha =-3,\alpha \beta \gamma =1\\ \therefore {\alpha }^{-2}+{\beta }^{-2}+{\gamma }^{-2}=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}\\ =\frac{{\left(\alpha \beta \right)}^2+{\left(\beta \gamma \right)}^2+{\left(\gamma \alpha \right)}^2}{{\left(\alpha \beta \gamma \right)}^2}\\ =\frac{{(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2\alpha \beta \gamma (\alpha +\beta +\gamma )}{{\left(\alpha \beta \gamma \right)}^2}\\ =\frac{9-2\left(1\right)(-2)}{1}\\ =13\end{array}$