# 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 .

• Your method looks good... Dec 8 '15 at 11:38

Another way:

$$\sum\dfrac1{r^2}=\dfrac{\sum s^2t^2}{(rst)^2}$$

Now $\sum s^2t^2=(\sum st)^2-rst(r+s+t)$

Now use Vieta's formula

Another approach:

Equation whose roots are squares of the roots of $$f(x)=x^3-6x^2+5x-7=0$$ is

$$f(\sqrt{x})=0$$ i.e., $$x\sqrt{x}+5\sqrt{x}=6x+7$$ squaring both sides and re arranging we get $$x^3-26x^2-59x-49=0$$ and equation whose roots are reciprocals to that of above is $$49x^3+59x^2+26x-1=0$$ and whose sum of the roots is $\frac{-59}{49}$

Let $y=\dfrac1{x^2}$

$$0=x^3-6x^2+5x-7=\dfrac{x-6}y+5x-7$$

$$\iff5x-7=\dfrac{6-x}y\iff 5xy-7y=6-x$$

$$\iff x(5y+1)=7y+6\iff x=\dfrac{7y+6}{5y+1}$$

Put this value in $$0=x^3-6x^2+5x-7$$ to form a cubic equation in $y$

Now use Vieta's formula

• Thanks for you answer (+1) .Is my work worth of a look at least ? Dec 8 '15 at 11:17
• @Mr.Y, If I transform the equation, I should use this method. Else I will use the method in my other answer Dec 8 '15 at 11:20

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: $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}\,\begin{pmatrix}-\frac{17}{49} & \frac{5}{7} & 1 \\ -\frac{23}{49} & -\frac{6}{7} & 0 \\ \frac{5}{49} & \frac{1}{7} & 0\end{pmatrix} = \color{red}{-\frac{59}{49}}.$$