$\displaystyle \alpha, \beta$ are zeroes of the polynomial $\displaystyle \mathrm{p}(x)=3 x^{2}-6 x-5$. Find the value of $\displaystyle \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$.

\[\begin{aligned} & \alpha+\beta=2, \alpha \beta=-\frac{5}{3} \\ & \therefore \frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{\alpha^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}=\frac{4+\dfrac{10}{3}}{\dfrac{25}{9}} \\ & =\frac{66}{25} \end{aligned} \]