Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?

Clearly, the two of the three roots of $$\frac1{a+x}+\frac1{b+x}+\frac1{c+x}=\frac2x$$ are $\omega,\omega^2$
On simplification, $$2x^3+2(a+b+c)x^2+2(ab+bc+ca)x+2abc=3x^3+2x^2(a+b+c)+x(ab+bc+ca)$$
$$\iff x^3+x^2(0)-(ab+bc+ca)x-2abc=0$$
If $a$ is the third root, using Veita's formula $$a+\omega^2+\omega=-\dfrac01$$
Hope the rest should be easy to deal with.
