If $$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac2w$$ where $a,b,c,d$ are real numbers and $w$ is a non real complex cube root of unity then prove that $$\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}+\frac{1}{d+w^2}=2w$$
I have no idea as how to start the problem. Please help me.
Hint:
$ω^2$ is the conjugate of $ω$, so $$\frac{1}{a+ω^2}+\frac{1}{b+ω^2}+\frac{1}{c+ω^2}+\frac{1}{d+ω^2}=\overline{\,\frac{1}{a+ω}+\frac{1}{b+ω}+\frac{1}{c+ω}+\frac{1}{d+ω}\,}.$$