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.
 A: 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,
$$\begin{align*}
r+s+t&=6 \\
rs+st+tr&=5 \\
rst&=7
\end{align*}$$
A: $$x^3-6x^2+5x-7=0$$
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, $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}\\=y_1^2+y_2^2+y_3^2\\=(y_1+y_2+y_3)^2-2(y_1y_2+y_2y_3+y_3y_1)\\=\left(\frac57\right)^2-2\left(\frac67\right)\\=-\frac{59}{49}.$$
A: You can also use this theorem:
Let $\alpha_1,\alpha_2, \alpha_3\cdot\cdot\cdot\alpha_n $ be the root of the polynomial equation $f(x)=0$ of the nth degree. Then the symmetric function:
$S_r=\alpha_1^{-r}+\alpha_2^{-r}+\cdot\cdot\cdot+\alpha_n^{-r}$
where r is a positive integer, is equal to the coefficient of $x^{r-1}$ in the expansion of $-f'(x)/f(x)$ in the  ascending powers of x, where $f'(x)$ is the first derived function.
For your equation we have:
$f'(x)=3x^2-12x +5$
$\frac{-f'(x)}{f(x)}=\frac{-5+12x-3x^2}{-7+5x-6x^2+x^3}=\frac 57-\frac {59}{47}x+\cdot\cdot\cdot$
You can find the expansion by direct division of $f'(x)$ by $f(x)$. So we have:
$\frac 1{r^2}+\frac1{s^2}+\frac 1{t^2}=-\frac {59}{49}$
A: Hint: Yet another way, is to transform the reciprocal polynomial $p(x)$ that you got in (1) to a polynomial $q(x)$ whose roots are squares of roots of $p(x)$. For this, $q(x^2)=p(x)p(-x)$ does the job (and you need only the first two coefficients for the sum sought).
