If $\omega$ is complex cube root of unity and $a,b,c$ are 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 the value of $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ is equal to ______________

$\because \frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2=\frac{2}{\omega }$ And $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}=2\omega =\frac{1}{{\omega }^2}$ 
It is clear that, $\omega$ and ${\omega }^2$ are the roots of the equation $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$ ……… (i)
$\Rightarrow x\Sigma (b+x)(c+x)=2(a+x)(b+x)(c+x)$ 
$\Rightarrow x^3-(ab+bc+ca)x-2abc=0$ 
$\therefore \alpha =1$ 
$\therefore$ Third root is 1.
From Eq. (i), we get $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$