Sum of complex roots' fractions According to this:
If $\omega^7 =1$ and $\omega \neq 1$ then find value of

$\displaystyle\frac{1}{(\omega+1)^2} +
\frac{1}{(\omega^2+1)^2} +
\frac{1}{(\omega^3+1)^2} +
... + \frac{1}{(\omega^6+1)^2}=?$

First I try like 
$\displaystyle\frac{1}{\omega+1} +
\frac{1}{\omega^2+1} +
\frac{1}{\omega^3+1} +
... + \frac{1}{\omega^6+1} = 3
$


I have done distribution them and finally got the solution $\dfrac{5}{3}$

 However, this is, without a doubt, a time-consuming way.
 Can someone please suggest easier way to solve this one.

Using $\omega^7=1$ the second sum computes to
$$
\frac{3(\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1)}{\omega^6 + 2\omega^4 + \omega^3 + \omega^2 + \omega + 1}=3
$$
For the first sum I do not obtain $5/3$. I obtain
$$
3\cdot\frac{5\omega^6+ 2\omega^5 + 11\omega^4 - 4\omega^3 + 11\omega^2 + 2\omega + 5}{9\omega^6 + 10\omega^5 + 7\omega^4 + 12\omega^3 + 7\omega^2 + 10\omega + 9}=-9,
$$
because $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$.
