Vrftsjtryk

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

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

This is as far as I could go...

Hints and full answers are welcomed.

$1$ 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
$$frac1{z^{n+1}-1}=frac a{z-1}+sum_{k=1}^nfrac{b_k}{x_kz-1}.$$
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
$$sum_{k=1}^nfrac1{x_kz-1}=frac{n+1}{z^{n+1}-1}-frac1{z-1}.$$
Letting $zto1$ gives
$$sum_{k=1}^nfrac1{x_k-1}=lim_{zto1}frac{n-z-z^2-cdots-z^n}{z^{n+1}-1}
=frac{-1-2-cdots-n}{n+1}=-frac n2.$$

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
$$S=sum_{k=1}^nfrac1{x_k^{-1}-1}=sum_{k=1}^nfrac{x_k}{1-x_k}
$$
and
$$2S=sum_{k=1}^nleft(frac1{x_k-1}+frac{x_k}{1-x_k}right)
=sum_{k=1}^n(-1)=-n.$$

Applying the same technique as per this answer
$$Q(x)=prodlimits_{k=1}^n(x-x_k) text{ and }
Q'(x)=sumlimits_{i=1}^{n}prodlimits_{k=1,kne i}^{n}left(x-x_kright)$$
then
$$frac{Q'(x)}{Q(x)}=sumlimits_{i=1}^{n}frac{1}{x-x_i}$$
or
$$-frac{Q'(1)}{Q(1)}=sumlimits_{i=1}^{n}frac{1}{x_i-1}$$
But $Q(1)=n+1$ and $Q'(x)=1+2x+3x^3+...+nx^{n-1}Rightarrow Q'(1)=frac{n(n+1)}{2}$ and
$$sumlimits_{i=1}^{n}frac{1}{x_i-1}=-frac{n}{2}$$

We have $$x^{n+1}-1=0 (1)$$ with $xne1$

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

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

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

Now apply Vieta's formula to find $$sum_{r=1}^ndfrac1{x_r-1}=sum_{r=1}^ny_r=-dfrac{binom{n+1}2}{binom{n+1}1}=?$$

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
$$
(x+1)^{n+1} -1 = xsum_{j=0}^{n} binom{n+1}{j+1}x^j = 0.
$$
Hence, $z_i = x_i -1$, $1leq i leq n$ are roots of
$$
sum_{j=0}^{n} binom{n+1}{j+1}z^j = 0,
$$ and by Vieta's formula,
$$
sum_{i=1}^n frac{1}{z_i} = frac{sum_{i=1}^n z_1z_2cdots z_{i-1}widehat{z_i}z_{i+1}cdots z_n}{z_1z_2cdots z_{n-1}z_n} = -frac{binom{n+1}{2}}{binom{n+1}{1}} = -frac{n}{2},
$$ where $widehat{z_i}$ means the variable is omitted in the calculation.

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