Value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ 
Given
$$
\tag1\frac{1}{\omega+a}+\frac{1}{\omega+b}+\frac{1}{\omega+c} = 2\omega^2
$$
and
$$
\tag2\frac{1}{\omega^2+a}+\frac{1}{\omega^2+b}+\frac{1}{\omega^2+c} = 2\omega
$$
what is the value of $\displaystyle \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$, where $\omega$ is a complex cube root of unity?

My Attempt:
Multiply $(1)$ by $\omega$ to get
$$
\frac{\omega}{\omega+a}+\frac{\omega}{\omega+b}+\frac{\omega}{\omega+c} = 2\omega^3 = 2
$$
(because $\omega^3 = 1$).
After simplification , we get
$$
\tag3\frac{a}{\omega+a}+\frac{b}{\omega+b}+\frac{c}{\omega+c} = 1
$$
Now multiply $(2)$ by $\omega^2$ to get
$$
\frac{\omega^2}{\omega^2+a}+\frac{\omega^2}{\omega^2+b}+\frac{\omega^2}{\omega^2+c} = 2\omega^3 = 2
$$
After simplification , we get
$$
\tag4\frac{a}{\omega^2+a}+\frac{b}{\omega^2+b}+\frac{c}{\omega^2+c} = 1
$$
From $(3)$ and $(4)$ we can form a quadratic equation whose roots are $x\in\left\{\omega,\omega^2\right\}$ as
$$
\frac{a}{x+a}+\frac{b}{x+b}+\frac{c}{x+c} = 1\\
\begin{align}
a\left[x^2+(b+c)x+bc\right]+b\left[x^2+(a+c)x+ac\right]&+c\left[x^2+(a+b)x+ab\right]\\
&=(x+a)(x+b)(x+c)\\
(a+b+c)x^2+2(ab+bc+ca)x+3abc&=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc\\
x^3-(ab+bc+ca)x-2abc&=0
\end{align}
$$
How can I solve the problem from this point?

This turned out to be a cubic.  There are three roots.  Two of them are $x=\omega$ and $x=\omega^2$.
HINT: Do you know how to find the sum of the roots of a polynomial, without solving the polynomial itself?
