<s>[INST] Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
how can I deduce
$$
\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1}?
$$
I have tried clearing the denominators of the first equation, but that just results in a large mess. I don't know how to continue from there.
 [/INST] The keys is Vieta's formulas.
Let $u$ be a root of $x^3 = 1 $. Then $u+a$ is a root of $(x-a)^3 = 1$ and $\frac{1}{u+a}$ is a root of $(\frac{1}{x}-a)^3=1 $. Multiplying by $x^3$ and expanding the coefficients we have
$$(1+a^3) x^3 -3a^2x^2 + 3ax - 1 = 0$$
Using  Vieta's formulas  we get that $\frac{3a^2}{1+a^3}$ is equal to
$$ \frac{3a^2}{1+a^3} = \sum_{u^3 =1} \frac{1}{(u+a)} \tag{1} $$
We can apply this to our equation. We know that
$$\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega}$$ and by conjugating: $$\frac{1}{a + \overline{\omega}} + \frac{1}{b +\overline{\omega}} + \frac{1}{c +\overline{\omega}} + \frac{1}{d + \overline{\omega}} = \frac{2}{\overline{\omega}}$$
Now $\omega$ and $\overline{\omega}$ are two of the three roots of unity (solutions to $x^3 = 1$), the other being $1$. From this we conclude:
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3}  + \frac{3d^2}{1+d^3}  - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3}  + \frac{3d^2}{1+d^3})    $$
</s>