If $\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}$, where $a,b,c,d\in\mathcal{R}$ and $\omega$ is a non-real cube root of unity, then prove that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$
As it was asked as a multiple choice question(this expression was one option to pick) I really do not see an easy way to prove this. $$ \frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\ \frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\ \frac{1}{a\omega^2+\omega}+\frac{1}{b\omega^2+\omega}+\frac{1}{c\omega^2+\omega}+\frac{1}{d\omega^2+\omega}=2\\ \frac{1}{a\omega+1}+\frac{1}{b\omega+1}+\frac{1}{c\omega+1}+\frac{1}{d\omega+1}=2\\ $$
Is it no coincidence that replacing $\omega$ with other two cube root of unity satisfy the equation ?