if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then... 
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
  $$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$

Now this has the solution too, but I do not understand the last step of the solution so here is the solution from the book.
$1,\, \alpha_1, \, \alpha_2,\, \ldots, \, \alpha_n$ are the $n^\text{th}$ of unity. These are the roots of $x^n-1=0$
Let $y=\frac{1}{1-\alpha}$ where $\alpha = \alpha_1, \, \alpha_2, \, \alpha_3, \, \ldots,  \, \alpha_n$ 
$$1 - \alpha = \frac{1}{y} \Rightarrow \alpha = \frac{y-1}{y}.$$
But $\alpha$ is a root of $x^n-1=0 \therefore \alpha^n=1 \Rightarrow (y-1)^n = y^n$ 
$$\Rightarrow y^n - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = y^n \\
\Rightarrow  - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = 0.$$
Sum of roots 
$$\frac{1}{1-\alpha_1}+\frac{1}{1-\alpha_2}+\ldots+\frac{1}{1-\alpha_{n-1}} = \frac{_nC_2}{_nC_1} = \frac{n-1}{2}$$
So this last part from "Sum of roots" I do not understand. I cannot see how this last shape relates to this binomial theorem notation. Can anyone help?
 A: Consider the polynomial
$$P(z)=-{_nC_1} z^{n-1}+{_nC_2}z^{n-2}-\cdots +(-1)^n.\tag{1}\label{1}$$
The answer shows that the numbers $y_i=\frac{1}{1-\alpha_i}$ are all roots of $P$, so they must be all the roots. Thus we must have
$$P(z)=-{_nC_1}(z-y_1)\cdots(z-y_{n-1})\tag{2}\label{2}$$
Then comparing the coefficients of $z^{n-2}$ in $P$ from \eqref{1} and \eqref{2}, we find
$${_nC_2}={_nC_1}\left(\sum_i y_i\right)$$
and thus
$$\sum_i y_i=\frac{_nC_2}{_nC_1}.$$
A: Since you already obtained some clarification on the textbook's solution, I am introducing a different solution.  You can also prove by noting that the roots of $x^n-1=0$ are $1,\xi,\xi^2,\ldots,\xi^{n-1}$, where
$$\xi=e^{\frac{2\pi i}{n}}.$$
Therefore, we may take $a_k$ to be $\xi^k$ for $k=1,2,\ldots,n-1$.  Now,
$$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\xi^{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{\xi^n}{\xi^k}}.$$
Since $\xi^n=1$, we obtain
$$\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}=\frac{1}{1-\xi^k}+\frac{1}{1-\frac{1}{\xi^k}}=\frac{1}{1-\xi^k}+\frac{\xi^k}{\xi^k-1}=1.$$
Therefore,
$$2\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\sum_{k=1}^{n-1}\left(\frac{1}{1-a_k}+\frac{1}{1-a_{n-k}}\right)=\sum_{k=1}^{n-1}1=n-1,$$
so
$$\sum_{k=1}^{n-1}\frac{1}{1-a_k}=\frac{n-1}{2}.$$
A: The sum of the roots of $a_nx^{n}+a_{n-1}x^{n-1}+\cdots +a_0$ is $-\frac {a_{n-1}} {a_n}$.
