If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to? If $ \alpha_i,  i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to?
A) $ \frac{n}{3^n-1} $
B) $ \frac{n-1}{3^n-1} $
C) $ \frac{n+1}{3^n-1} $
D) $ \frac{n+2}{3^n-1} $
Attempt: I know that $(3- \alpha_0)(3- \alpha_1)....(3-\alpha_{n-1})= (3^n-1)/2 $
But I have no clue about the numerator. Adding 3 and subracting 3 from the numberator would make the fraction simpler, but I would still have to sum up $ 1/(3- \alpha_i) $

Let $\dfrac{a_i}{3-a_i}=b_i\iff a_i=\dfrac{3b_i}{1+b_i}$
As $a_i^n=1,$
$$\left(\dfrac{3b_i}{1+b_i}\right)^n=1\iff(3^n-1)b_i^n-\binom n1b_i^{n-1}+\cdots=0$$
By Vieta's formula,
$$\displaystyle\sum_{i=0}^{n-1}b_i=\dfrac{\binom n1}{3^n-1}$$
