<s>[INST] the sum of the reciprocal of the squares of roots $$x^3-6x^2+5x-7=0$$
Find $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$ where $r,s,t$ are roots of the equation.

*

*First I got the reciprocal of the equation above and I got $-7x^3+5x^2-6x+1=0$

*Using newtons identity/sum the sum of all the roots of this equation is $5/7$. The sum of the squares of the equations can be found by solving  $-7s+5(5/7)-12=0$, for $s$ (s is the sum of the squares of the roots). I get $59/-49$
Question: Am I correct or are my steps wrong. I know that if you simplify $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$  you get $\frac{s^2t^2+r^2t^2+r^2s^2}{r^2s^2t^2}$ but I don't know how to calculate the numerator of the fraction.
Thanks in advance.
 [/INST] Hint:
$$\begin{align*} \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}&=\frac{r^2s^2+s^2t^2+t^2r^2}{r^2s^2t^2} \\ \\
(rs+st+tr)^2&=r^2s^2+s^2t^2+t^2r^2+2rst(r+s+t).
\end{align*}$$
Also,
$$\begin{align*}
r+s+t&=6 \\
rs+st+tr&=5 \\
rst&=7
\end{align*}$$
</s>