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.

Let $\dfrac1{w+1}=x\implies w=\dfrac{1-x}x$
$$\implies\left(\dfrac{1-x}x\right)^7=1$$
As $x\ne0,$ $$x^6-3x^5+5x^4-\cdots=0$$
We need $$\sum_{r=1}^6x_r^2=\left(\sum_{r=1}^6x_r\right)^2-2\sum_{1\le i< j\le6}x_ix_j=\left(\dfrac31\right)^2-2\cdot\dfrac51$$
