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 $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}$$
 A: 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 $\alpha^2\beta^2\gamma^2 = 3^2$. 
A: 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:
$$ \frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=-\left.\frac{d^2}{dx^2}\log\left(1-2x+\frac{5}{3}x^2-\frac{1}{3}x^3\right)\right|_{x=0}=\color{red}{\frac{2}{3}}.\tag{4}$$
A: $$T:=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\left(\frac1\alpha+\frac1\beta+\frac1\gamma\right)^2-2\left(\frac1{\alpha\beta}+\frac1{\alpha\gamma}+\frac1{\beta\gamma}\right)$$
and
$$\frac1\alpha+\frac1\beta+\frac1\gamma=\frac{\alpha\beta+\alpha\gamma+\beta\gamma}{\alpha\beta\gamma}=\frac63=2$$
and
$$\frac1{\alpha\beta}+\frac1{\alpha\gamma}+\frac1{\beta\gamma}=\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}=\frac53$$
so finally
$$T=4-\frac{10}3=\color{red}{\frac23}$$
A: Put $u_n=\alpha^n+\beta^n+\gamma^n$. Then we have $$u_{n+3}-5u_{n+2}+6u_{n+1}-3u_n=0$$ for all $n\in \mathbb{Z}$. Now $u_1=\alpha+\beta+\gamma=5$, $u_0=3$, $u_{-1}=(\alpha\beta+\beta\gamma+\gamma\alpha)/\alpha\beta\gamma=2$. Putting $n=-2$ in the relation above gives $u_{-2}=\frac{2}{3}$. 
