341\sum \frac{1}{a+\omega} = 2 \omega^2 \Rightarrow \sum \frac {\omega}{a+\omega} = 2 \Rightarrow \sum \frac {a}{a+\omega} = 1
Similarly with the second relation we get \sum \frac {a}{a+\omega^2} = 1
Hence if we consider \sum \frac {a}{a+x} = 1 , which simplifies to x^3 - \left(\sum ab \right) \ x - 2abc = 0, we already know \omega, \omega^2 to be roots and the sum of roots to be zero, the third root must be 1.
Hence we have \sum \frac{a}{a+1} = 1 \Rightarrow 3 - \sum \frac {1} {a+1} =1 \Rightarrow \sum \frac{1}{a+1} = 2