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 $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$$

Given
$$ x^3-5x^2+6x-3 = (x-\alpha)(x-\beta)(x-\gamma)\tag{1} $$
we have $\alpha\beta\gamma=3$, hence:
$$ \left(1-\frac{x}{\alpha}\right)\left(1-\frac{x}{\beta}\right)\left(1-\frac{x}{\gamma}\right) = 1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3 \tag{2}$$
and by switching to logarithms:
$$ -\sum_{n\geq 1}\sum_{\xi\in\{\alpha,\beta,\gamma\}}\frac{x^n}{n\xi^n} = \log\left(1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3\right)\tag{3} $$
hence:
$$ \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=-\left.\frac{d^2}{dx^2}\log\left(1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3\right)\right|_{x=0}=\color{red}{\frac{2}{3}}.\tag{4}$$
