How complex can it get

Let $a$ , $b$ , $c$ and $d$ be $a_1$ , $a_2$ , $a_3$ and $a_4$ respectively. Therefore,

\(\begin{array} {} \displaystyle \sum_{k=1}^4 \frac{1}{a_k + \omega} & = \displaystyle \sum_{k=1}^4 \frac{a_k + \omega^2}{(a_k + \omega)(a_k + \omega^2)} \\ & = \displaystyle \sum_{k=1}^4 \frac{a_k + \omega^2}{a_k^2 + (\color{red}{\omega + \omega^2})a_k + \color{blue}{\omega^3}} & \small \color{red} {[\omega^2 + \omega + 1 = 0]} \\ & = \displaystyle \sum_{k=1}^4 \frac{a_k + \omega^2}{a_k^2 \color{red}{-}a_k + \color{blue}{1}} & \small \color{blue} {[\omega^3 = 1 ]} \end{array} \)

$\begin{aligned} \Rightarrow \sum_{k=1}^4 \frac{\color{#3D99F6}{a_k} + \color{#D61F06}{\omega^2}}{a_k^2 - a_k + 1} & = \dfrac{2}{\omega} = \frac{2\omega^3}{\omega} = \color{#3D99F6}{0} + \color{#D61F06}{2\omega^2} \end{aligned}$

$\begin{aligned} \Rightarrow \color{#3D99F6} {\sum_{k=1}^4 \frac{a_k}{a_k^2 - a_k + 1}} + \color{#D61F06} {\omega^2 \sum_{k=1}^4 \frac{1}{a_k^2 - a_k + 1}} & = \color{#3D99F6}{0} + \color{#D61F06}{2\omega^2} \end{aligned}$

$\Rightarrow \begin{cases} \displaystyle \color{#3D99F6} {\sum_{k=1}^4 \frac{a_k}{a_k^2 - a_k + 1} = 0} \\ \displaystyle \color{#D61F06} {\sum_{k=1}^4 \frac{1}{a_k^2 - a_k + 1}=2} \end{cases}$

Similarly,

$\begin{aligned} \sum_{k=1}^4 \frac{1}{a_k + \omega^2} & = \sum_{k=1}^4 \frac{\color{#3D99F6}{a_k} + \color{#D61F06}{\omega}}{a_k^2 - a_k + 1} = \color{#D61F06}{2\omega} = \frac{2\omega}{\omega^3} = \boxed{\dfrac{2}{\omega^2}}\end{aligned}$

Simply take complex conjugate on both sides . and use the fact that a,b,c,d are real numbers so there complex conjugate will be a,b,c,d itself . also both the complex cube roots of unity are also the complex conjugate of each other . thats it!

please can somebody post a solution i was not able to solve this!