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

Question

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: $\sum^{20}_{k=1}\frac{1}{1-x_k}$.

Answer 1

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

Answer 2

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?