# Calculate $\sum^{20}_{k=1}\frac{1}{x_k-x_k^2}$ where $x_k$ are roots of $P(x)=x^{20}+x^{10}+x^5+2$

We have the polynomial $$P(x)=x^{20}+x^{10}+x^5+2$$, which has roots $$x_1,x_2,x_3,...,x_{20}$$. Calculate the sum $$\sum^{20}_{k=1}\frac{1}{x_k-x_k^2}$$

What I've noticed: $$\sum^{20}_{k=1}\frac{1}{x_k-x_k^2}=\sum^{20}_{k=1}\left(\frac{1}{x_k}+\frac{1}{1-x_k}\right)$$

I know how to calculate the first sum: $$\sum^{20}_{k=1}\frac{1}{x_k}$$.

Please help me calculate the second one: $$\sum^{20}_{k=1}\frac{1}{1-x_k}$$.

• If $x_k$ are roots of $P(x)$, then $y_k=1-x_k$ are roots of $P(1-x)$. Maybe that can help? – Sil Mar 22 at 19:38
• Hint: $\frac{P'(x)}{P(x)} = \sum_{k=1}^{20}\frac{1}{x-x_k}$ – achille hui Mar 22 at 19:44
• Some of your tags have no connection with the issue : linear-algebra, symmetric polynomials (we don't have a symmetry of coefficients), abstract algebra (no group, ring, field, etc... here). I have taken the liberty to suppress them. I have added the "roots" tags. – Jean Marie Mar 23 at 16:11

Since $$\frac{P'(x)}{P(x)} = \sum_{k=1}^{20}\frac{1}{x-x_k}$$

and $$P'(x)= 20x^{19}+10x^9+5x^4$$

we have $$\sum_{k=1}^{20}\frac{1}{1-x_k}=\frac{P'(1)}{P(1)} = {35\over 5}=7$$

• Why is $$\frac{P'(x)}{P(x)} = \sum_{k=1}^{20}\frac{1}{x-x_k}$$ – Dr. Mathva Mar 24 at 9:34
• Write $P(x)= a_n(x-x_1)(x-x_2)...$ take it log and then derviate it@Dr.Mathva – Maria Mazur Mar 24 at 11:02

Hint:

Set $$y=1-x$$. If the $$x_k$$ satisfy the equation $$\;x^{20}+x^{10}+x^{5}+2=0$$, the corresponding $$\:y_k$$ satisfy the equation $$(1-y)^{20}+(1-y)^{10}+(1-y)^{5}+2=0.$$

Can you find the constant term and the coefficient of $$y$$ in this equation, to use Vieta's relations?