<s>[INST] find $ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$ given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$ 
I am asked to find $$ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$$
  given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$ .

So what I did was to get the polynomial whose roots are the reciprocals of $r,s,t$ ,namely $$-7x^3+5x^2-6x+1=0$$ 
From that, I've then used the $2^{nd}$ Newton'sum  $a_ns_2+a_{n-1}s_1+2a_{n-2}=0$.
Now given that $s_1=- \cfrac{a_{2}}{a_3}$ I have in my case $s_1=\cfrac{5}{7}$ (I've turned my polynomial into a monic polynomial) .
Therefore my Newton's sum is  $$s_2-\cfrac{5}{7}s_1+2\cdot\cfrac{6}{7}=0$$ 
$$s_2-\cfrac{5}{7}\cdot\cfrac{5}{7} +\cfrac{12}{7}=0$$
Solving for $s_2$ I get $s_2=-\cfrac{59}{49}\approx -1,204  $
Now ,I've just started studying this stuff so I am not sure  I've applied it all right(the hardest part for me is to get the signs right),so I would be really grateful if someone could verify my work .
Thanks in advance.
 [/INST] Still another approach. We have that
$$ M=\begin{pmatrix}0 & 0 & 7 \\ 1 & 0 & -5 \\ 0 & 1 & 6\end{pmatrix}$$
is the companion matrix of your polynomial, and:
$$ \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}(M^{-2})=\text{Tr}\,\begin{pmatrix}0 & 7 & 42 \\ 0 & -5 & -23 \\ 1 & 6 & 31\end{pmatrix}^{-1} $$
hence:
$$ \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}\,\begin{pmatrix}-\frac{17}{49} & \frac{5}{7} & 1 \\ -\frac{23}{49} & -\frac{6}{7} & 0 \\ \frac{5}{49} & \frac{1}{7} & 0\end{pmatrix} = \color{red}{-\frac{59}{49}}.$$
</s>