If $\omega$ is complex cube root of unity and a, b, c are three real numbers such that $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2$ and $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega$ , then$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=$  

( Since ${\omega }^2=\frac{1}{\omega }$ and $\omega =\frac{1}{{\omega }^2}$  )

the given relation may be rewritten as $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=\frac{2}{\omega }$

$and\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=\frac{2}{{\omega }^2}$

Clearly $\omega$  and ${\omega }^2$  are the roots of

$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$  or $\frac{\left(b+x\right)\left(c+x\right)+\left(a+x\right)\left(c+x\right)+\left(a+x\right)\left(b+x\right)}{\left(a+x\right)\left(b+x\right)\left(c+x\right)}=\frac{2}{x}$

$orx\left[3x^2+2\left(a+b+c\right)x+bc+ca+ab\right]$

$=2\left[abc+\left(bc+ca+ab\right)x\left(a+b+c\right)x^2+x^3\right]$

$\Rightarrow x^3-\left(bc+ca+ab\right)x-2abc=0$

Now if $\alpha$ is the third root of this equation then sum of the roots,$\alpha +\omega +{\omega }^2=0\Rightarrow \alpha =1$

Hence,1 is the root of equation(1) we get $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$