<s>[INST] 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?
 [/INST] I would like to add to the above answer, the symmetry put a spark in my brain and i came up with this.
As in answer stated by peiter.
Assume  
$X=\large\frac{1}{a+1}+\large\frac{1}{b+1}+\large\frac{1}{c+1}$ $\space\space\space\space\space\space\space\space$ and
$\large\frac{1}{a+\omega}+\large\frac{1}{b+\omega}+\large\frac{1}{c+\omega}=2\omega^2$
$\large\frac{1}{a+\omega^2}+\large\frac{1}{b+\omega^2}+\large\frac{1}{c+\omega^2}=2\omega$  
Adding the three equation and considering first term related to "$a$"
$X+2\omega+2\omega^2=\large\frac{1}{a+1}+\large\frac{1}{a+\omega}+\large\frac{1}{a+\omega^2}+other\space terms$  
simplifies to,  
$X+2\omega+2\omega^2=\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$  
Now as according to solution given by pieter $X=2$  
Therefore  
$\large\frac{3a^2}{a^3+1}+\large\frac{3b^2}{b^3+1}+\large\frac{3c^2}{c^3+1}$   must equal zero.  
If somehow you can prove that , the answer may become simpler.(i.e., to prove $X=2$)  
I would suggest trying to multiplying and dividing the express by $\omega$ and $\omega^2$ to arrive at other expressions that may yield this result.
</s>