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Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$

Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)

My attempt: Developing expression $P(1)$, replacing the 1 by $x$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$

As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that

$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $P(1)$ is $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.

Another fact that is probably useful is that $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is

$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$

This is as far as I could go...

Hints and full answers are welcomed.

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$1$ and $x_1,\ldots,x_n$ are the roots of $z^{n+1}-1=0$. They are the $(n+1)$-th roots of unity. Then $1/(z^{n+1}-1)$ has a partial fraction expansion of the form $$\frac1{z^{n+1}-1}=\frac a{z-1}+\sum_{k=1}^n\frac{b_k}{x_kz-1}.$$ Multiplying by $z-1$ and setting $z=1$ gives $a=1/(n+1)$. Multiplying by $x_kz-1$ and setting $z=1/x_k$ gives $b_k=1/(n+1)$. Therefore $$\sum_{k=1}^n\frac1{x_kz-1}=\frac{n+1}{z^{n+1}-1}-\frac1{z-1}.$$ Letting $z\to1$ gives $$\sum_{k=1}^n\frac1{x_k-1}=\lim_{z\to1}\frac{n-z-z^2-\cdots-z^n}{z^{n+1}-1} =\frac{-1-2-\cdots-n}{n+1}=-\frac n2.$$

ADDED IN EDIT.

Here's a trick proof. Let $S$ denote the sum in question. As the reciprocals of the $x_k$ are the $x_k$ again (in a different order) then $$S=\sum_{k=1}^n\frac1{x_k^{-1}-1}=\sum_{k=1}^n\frac{x_k}{1-x_k} $$ and $$2S=\sum_{k=1}^n\left(\frac1{x_k-1}+\frac{x_k}{1-x_k}\right) =\sum_{k=1}^n(-1)=-n.$$

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Applying the same technique as per this answer $$Q(x)=\prod\limits_{k=1}^n(x-x_k) \text{ and } Q'(x)=\sum\limits_{i=1}^{n}\prod\limits_{k=1,k\ne i}^{n}\left(x-x_k\right)$$ then $$\frac{Q'(x)}{Q(x)}=\sum\limits_{i=1}^{n}\frac{1}{x-x_i}$$ or $$-\frac{Q'(1)}{Q(1)}=\sum\limits_{i=1}^{n}\frac{1}{x_i-1}$$ But $Q(1)=n+1$ and $Q'(x)=1+2x+3x^3+...+nx^{n-1}\Rightarrow Q'(1)=\frac{n(n+1)}{2}$ and $$\sum\limits_{i=1}^{n}\frac{1}{x_i-1}=-\frac{n}{2}$$

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We have $$x^{n+1}-1=0\ \ \ \ (1)$$ with $x\ne1$

Set $\dfrac1{x-1}=y\iff x=\dfrac{y+1}y$

Replace the value of $x$ in terms of $y$ in $(1)$ to form an $n$ degree equation in $y$ $$\left(\dfrac{y+1}y\right)^{n+1}=1$$

$$\implies\binom{n+1}1 y^n+\binom{n+1}2 y^{n-1}+\cdots+1=0$$

Now apply Vieta's formula to find $$\sum_{r=1}^n\dfrac1{x_r-1}=\sum_{r=1}^ny_r=-\dfrac{\binom{n+1}2}{\binom{n+1}1}=?$$

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Simply note that $1,x_1,\ldots, x_n$ are roots of $$ x^{n+1} -1 = 0.$$ Thus we have $0,x_1-1,\ldots, x_n-1$ are roots of $$ (x+1)^{n+1} -1 = x\sum_{j=0}^{n} \binom{n+1}{j+1}x^j = 0. $$ Hence, $z_i = x_i -1$, $1\leq i \leq n$ are roots of $$ \sum_{j=0}^{n} \binom{n+1}{j+1}z^j = 0, $$ and by Vieta's formula, $$ \sum_{i=1}^n \frac{1}{z_i} = \frac{\sum_{i=1}^n z_1z_2\cdots z_{i-1}\widehat{z_i}z_{i+1}\cdots z_n}{z_1z_2\cdots z_{n-1}z_n} = -\frac{\binom{n+1}{2}}{\binom{n+1}{1}} = -\frac{n}{2}, $$ where $\widehat{z_i}$ means the variable is omitted in the calculation.

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First, the denominator is, I think, $(-1)^nQ(1)$. Here is a hint for the numerator: how can you write $Q’(x)$?

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