If $\mathrm{\omega }$ is complex cube root of unity and a,b,c are such that $\frac{1}{\mathrm{a}+\mathrm{\omega }}+\frac{1}{\mathrm{b}+\mathrm{\omega }}+\frac{1}{\mathrm{c}+\mathrm{\omega }}=2{\mathrm{\omega }}^2$ and $\frac{1}{\mathrm{a}+{\mathrm{\omega }}^2}+\frac{1}{\mathrm{b}+{\mathrm{\omega }}^2}+\frac{1}{\mathrm{c}+{\mathrm{\omega }}^2}=2\mathrm{\omega }$ then the value of  $\frac{1}{\mathrm{a}+1}+\frac{1}{\mathrm{b}+1}+\frac{1}{\mathrm{c}+1}$ is equal to

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