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.

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*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.

$$x^3-6x^2+5x-7=0$$
Let $y=\frac1x.$ Then $$\frac1{y^3}-\frac6{y^2}+\frac5{y}-7=0\\7y^3-5y^2+6y-1=0.$$
So, by Vieta's Formulae, $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}\\=y_1^2+y_2^2+y_3^2\\=(y_1+y_2+y_3)^2-2(y_1y_2+y_2y_3+y_3y_1)\\=\left(\frac57\right)^2-2\left(\frac67\right)\\=-\frac{59}{49}.$$
