<s>[INST] The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$. The cubic equation $x^3-5x^2+6x-3 = 0$ has solutions $\alpha$, $\beta$ and $\gamma$.
Find the value of $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$$
 [/INST] Since $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{\alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2\gamma^2}{\alpha^2 \beta^2 \gamma^2}$$
You know that for a cubic $x^3 + ax^2 + bx + c$ the sum of roots are given by $\alpha + \beta + \gamma = -a = 5$, the sum of the alternate pairs $\alpha\beta + \beta\gamma + \alpha \gamma = b = 6$ and the product $\alpha \beta \gamma = -c = 3$. 
So squaring the second equation gives us $$\alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 + 2\alpha\beta\gamma(\alpha + \beta +\gamma) = 6^2$$
This yields $\alpha^2\beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 = 36 -2(3)(5) = 6$ and hence $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{6}{3^2} = \frac{6}{9} = \color{red}{\frac{2}{3}}$$
since squaring the third equation gives us $\alpha^2\beta^2\gamma^2 = 3^2$. 
</s>