<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] My original answer did not use the nice symmetry. Kelenner gave a better answer. Based on that, I have made yet another answer, using essentially the same method as Kelenner did, but with a different definition of the same function which is (for me) easier to understand. When I first wrote this answer, I did not realize I was using exactly the same function, otherwise I would not have written this down, but now that it is here I'll leave it, because it might be easier to read for some people.
Define the function $Q$ by
$$Q(x)=\left(\left(\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}\right) x-2\right)(a+x)(b+x)(c+x)$$
From the given equations:
$$Q(\omega)=0,\quad Q(\omega^2)=0.$$
The function $Q$ can be much simplified (with simple manipulations):
$$Q(x)=x^3-2abc$$
Because $\omega$ and $\omega^2$ are zeroes of $Q$, $Q$ must have a factor $(x^2+x+1)$. Following symbolic division, this can only be true if $(1-2abc)=0$, and then
$$Q(x)=(x^2+x+1)*(x-1)$$
So, $Q(1)=0$, and from the original definition of $Q$ it must follow that
$$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=2$$.
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