<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] Put $A(x)=(x+a)(x+b)(x+c)=x^3+ux^2+vx+w$. Then
$$\frac{A^{\prime}(x)}{A(x)}=\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}$$ Hence we get $A^{\prime}(\omega)=2\omega^2 A(\omega)$ and $A^{\prime}(\omega^2)-2\omega A(\omega^2)=0$, or if $Q(x)=xA^{\prime}(x)-2A(x)$, that $Q(\omega)=Q(\omega^2)=0$. Hence $Q(x)$ is divisible by $(x-\omega)(x-\omega^2)=x^2+x+1$. We have
$$Q(x)=x^3-vx-2w=(x-1)(x^2+x+1)+(-vx+1-2w)$$
Hence $v=0$, $w=1/2$, and $Q(x)=(x-1)(x^2+x+1)$, $A'(1)=2A(1)$, and we are done.
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