<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] Multiply all equations by the product of the denominators ($(a-\omega)(b-\omega)(c-\omega)$ in (1), $(a-\omega^2)(b-\omega^2)(c-\omega^2)$ in (2) etc.), to get rid of the fractions.
The first two equations then simplify to (remember $\omega^3=1$):
$$ 0=(ab+ac+bc)+(2abc-1)\omega^2 $$
$$ 0=(ab+ac+bc)+(2abc-1)\omega $$
Because $\omega\neq\omega^2$, these can only both be true if $2abc-1=0$. In that case both equations become 
$$ab+ac+bc=0.$$
If I replace the questionmark in your equation (3) by $X$, the equation becomes (after multiplying with the three terms):
$$X=\frac{(b+1)(c+1)+(a+1)(c+1)+(a+1)(b+1)}{(a+1)(b+1)(c+1)}$$
Expanded:
$$X=\frac{ab+bc+ac+2a+2b+2c+3}{abc+ab+ac+bc+a+b+c+1}$$
Use the fact that $ab+ac+bc=0$ and $2abc-1=2$:
$$X=\frac{2a+2b+2c+3}{a+b+c+\frac{3}{2}}=2$$
There must be a nicer way to show this, considering the nice symmetry in the problem, but I can't see that now.
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