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.
 A: A calculation by foot. We obtain from
\begin{align*}
\color{blue}{0}&\color{blue}{=\frac{1}{a+\omega}+\frac{1}{b+\omega}
+\frac{1}{c+\omega}+\frac{1}{d+\omega}-\frac{2}{\omega}}\tag{1}\\
\end{align*}
by multiplication with the common denominator $(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega$:
\begin{align*}
\color{blue}{0}&=(b+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(c+\omega)\omega\\
&\qquad-2(a+\omega)(b+\omega)(c+\omega)(d+\omega)\\
&=bcd\omega+(bc+bd+cd)\omega^2+(b+c+d)\omega^3+\omega^4\\
&\qquad+acd\omega+(ac+ad+cd)\omega^2+(a+c+d)\omega^3+\omega^4\\
&\qquad+abd\omega+(ab+ad+bd)\omega^2+(a+b+d)\omega^3+\omega^4\\
&\qquad+abc\omega+(ab+ac+bc)\omega^2+(a+b+c)\omega^3+\omega^4\\
&\qquad-2abcd\\
&\qquad-2(abc+abd+acd+bcd)\omega\\
&\qquad-2(ab+ac+ad+bc+bd+cd)\omega^2\\
&\qquad-2(a+b+c+d)\omega^3\\
&\qquad-2\omega^4\\
&\,\,\color{blue}{=\left(2-(abc+abd+acd+bcd)\right)\omega-2abcd+a+b+c+d}\tag{2}
\end{align*}
In the last step (2) we observe that terms with $\omega^2$ cancel away and we also use the identities
\begin{align*}
\omega^3=1,\quad\omega^4=\omega
\end{align*}

We note from (1) and (2) we can write (1) as
\begin{align*}
\color{blue}{0=\frac{A\omega+B}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega}}\tag{3}
\end{align*}
and since $A,B\in\mathbb{R}$ and $\omega\in\mathbb{C}\setminus{\mathbb{R}}$ we conclude $A=B=0$, so that
\begin{align*}
\color{blue}{abc+abd+acd+bcd}&\color{blue}{=2}\tag{4}\\
\color{blue}{a+b+c+d}&\color{blue}{=2abcd}\\
\end{align*}
follows.

On the other hand we consider the expression
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}&\color{blue}{=z}
\end{align*}
Multiplication of the LHS with the common denominator $(1+a)(1+b)(1+c)(1+c)$ gives

\begin{align*}
&(b+1)(c+1)(c+1)+(a+1)(c+1)(d+1)\\
&\qquad\quad+(a+1)(b+1)(d+1)+(a+1)(b+1)(c+1)\\
&\quad=1+(b+c+d)+(bc+bd+cd)+bcd\\
&\qquad\quad+1+(a+c+d)+(ac+ad+cd)+acd\\
&\qquad\quad+1+(a+b+d)+(ab+ad+bd)+abd\\
&\qquad\quad+1+(a+b+c)+(ab+ac+bc)+abc\\
&\quad=4+3(a+b+c+d)\\
&\qquad\quad+2(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+abc+abd+acd+bcd\\
&\,\,\color{blue}{=2\left(3+3abcd+ab+ac+ad+bc+bd+cd\right)}\tag{5}
\end{align*}
In the last line (5) we used the identities from (4).

Similarly, multiplication of the RHS with the common denominator gives

\begin{align*}
&z(a+1)(b+1)(c+1)(d+1)\\
&\qquad=z(1+(a+b+c+d)\\
&\qquad\quad+(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+(abc+abd+acd+bcd)+abcd)\\
&\,\,\color{blue}{\qquad=z(3+3abcd+ab+ac+ad+bc+bd+cd)}\tag{6}
\end{align*}
Again in the last line (6) we used the identities from (4) for simplification.

Comparing (5) and (6) we conclude
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2}
\end{align*}
A: I will write $u$ instead of $\omega$, so that $u$ is the primitive third root of unity, and the other one is $u^2=\bar u$.
Below, $v$ will be an element in the list $L=\{1,u,u^2\}$ of all third roots of unity.
The given relation can be rewritten:
$$
2=u\sum\frac 1{a+u}=\sum\frac u{a+u}=\sum\frac 1{1+au^2}
\ ,
$$
where the above sums have $4$ terms each, they are obtained by substituting instead of $a$ formally the values $a,b,c,d$.
It is natural to consider now the polynomial expressions which are the numerators of
$\displaystyle
2-\sum\frac 1{1+av}
$ for $v\in L$. They are
$$
\begin{aligned}
P_v
&:=2(1+av)(1+bv)(1+cv)(1+dv) 
\\
&\qquad\qquad
-\sum \color{gray}{\underbrace{(1+av)}_{\text{omitted}}}(1+bv)(1+cv)(1+dv)
\\
&=2v\; abcd + (bcd+cda+dab+abc) - v(a+b+c+d) -2
\\
\\[3mm]
&\qquad\text{From here we get immediately:}\\[3mm]
vP_v &=
2v^2\; abcd + v(bcd+cda+dab+abc) - v^2(a+b+c+d) -2v
\\
\sum_{v\in L}v\;P_v
&=
2\left(\sum v^2\right)\; abcd + \left(\sum v\right)(bcd+cda+dab+abc)
\\
&\qquad\qquad - \left( \sum v^2\right)(a+b+c+d) -2\left(\sum v\right)
\\
&=0
\ .
\end{aligned}
$$
We have used
$\sum v=\sum v^2=1+u+u^2=0$ for $v$ running in $L$ in the above sums. From $P_u=0$, and its conjugated cousin $P_{u^2}=0$, we obtain $P_1=0$. So:
$$
2=\sum\frac 1{1+a}\ .
$$
$\square$
A: 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:
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1} = \\ = \frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3}  + \frac{3d^2}{1+d^3}  - (\frac{2}{\omega}+\frac{2}{\overline{\omega}}) =\\= 2 + (\frac{3a^2}{1+a^3} + \frac{3b^2}{1+b^3} + \frac{3c^2}{1+c^3}  + \frac{3d^2}{1+d^3})    $$
