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?

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,

$$\displaystyle\sum_{i=0}^{n-1}b_i=\dfrac{\binom n1}{3^n-1}$$

$$\frac{u}{3-u}=\frac{3}{3-u}-1$$

So your sum is:

$$-n+3\sum \frac{1}{3-a_i}$$

Letting $p(x)=x^n-1$, show that $\frac{p'(x)}{p(x)}=\sum \frac{1}{x-a_i}$.

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More generally, if $p(x)=(x-b_1)(x-b_2)\cdots (x-b_n)$ then $$\sum \frac{b_i}{b-b_i} = -n + b\frac{p'(b)}{p(b)}$$