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.
 A: Note that $\omega, \ldots, \omega^6$ are precisely the roots of the sixth degree polynomial:
$$p(x) = x^6 + \cdots + 1 = \dfrac{x^7-1}{x-1}.$$
Thus, we can write
$$p(x) = (x-\omega)\cdots(x-\omega^6).$$
Taking (natural) $\log$ on both sides and differentiating gives us
$$\dfrac{p'(x)}{p(x)} = \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$
Note that
\begin{align}
\log p(x) &= \log(x^7 - 1) - \log(x-1)\\
\implies \dfrac{p'(x)}{p(x)} &= \dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1}.
\end{align}
This gives us that
$$\dfrac{7x^6}{x^7-1} - \dfrac{1}{x-1} =  \dfrac{1}{x-\omega}+\cdots+\dfrac{1}{x-\omega^6}.$$
Differentating both sides again gives us
$$\dfrac{(x^7-1)(42x^5) - (7x^6)(7x^6)}{(x^7-1)^2} + \left(\dfrac{1}{x-1}\right)^2 = -\left(\dfrac{1}{x-\omega}\right)^2-\cdots-\left(\dfrac{1}{x-\omega^6}\right)^2.$$
Now, we simply substitute $x = -1$ both sides. It is clear that the RHS transforms to the negative of what we want, whereas the LHS becomes
\begin{align}
\dfrac{(-2)(-42) - (7)(7)}{(-2)^2} + \left(\dfrac{1}{-2}\right)^2 &= \dfrac{84-49}{4} + \dfrac{1}{4}\\
&= \dfrac{36}{4} = 9
\end{align}
This gives us the answer as $-9$.
A: 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 $$\sum_{r=1}^6x_r^2=\left(\sum_{r=1}^6x_r\right)^2-2\sum_{1\le i< j\le6}x_ix_j=\left(\dfrac31\right)^2-2\cdot\dfrac51$$
A: 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 $\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1=0$.
