<s>[INST] Polynomial sum relation Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)
 [/INST] Note that the roots of $x^2+x+1$ are 
$$\zeta_3 =e^{2\pi i/3} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$$
and 
$$\zeta_3^2 =e^{4\pi i/3} = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$$
Both of these are roots of $P(x)$ with multiplicity $100$. Therefore,
$$\sum_{k=1}^{200}\frac{1}{1+x_k} = 100\left( \frac{1}{1+\zeta_3}+\frac{1}{1+\zeta_3^2}\right)$$
$$ = 100\left(\frac{1+\zeta_3+1+\zeta_3^2}{1+\zeta_3+\zeta_3^2+\zeta_3^3} \right) = 100$$
As for the polynomial division, suppose
$$P(x) = g(x)(x^2+x)+r(x)$$
where $r$ is linear. We have $P(0) = 1$, which implies that $r(0) = 1$. Further, $P(-1) = 1$, and so $r(-1)=1$. Consequently, $r(x)=1$ for all $x$, so $a_1=0$.
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