$$\begin{gathered} \mathrm{lf} \mathrm{\omega }\ne 1\mathrm{is} \mathrm{a} \mathrm{cube} \mathrm{root} \mathrm{of} \mathrm{unity} \mathrm{satisfying} \\ \frac{1}{\mathrm{a}+\mathrm{w}}+\frac{1}{\mathrm{b}+\mathrm{w}}+\frac{1}{\mathrm{c}+\mathrm{w}}=2{\mathrm{w}}^2 \mathrm{and} \\ \frac{1}{\mathrm{a}+{\mathrm{w}}^2}+\frac{1}{\mathrm{b}+{\mathrm{w}}^2}+\frac{1}{\mathrm{c}+{\mathrm{w}}^2}=2\mathrm{w} \mathrm{then} \mathrm{the} \\ \mathrm{value} \mathrm{of} \frac{1}{\mathrm{a}+1}+\frac{1}{\mathrm{b}+1}+\frac{1}{\mathrm{c}+1} \mathrm{is} \end{gathered}$$

$\begin{array}{rcl}\frac{1}{\mathrm{a}+\mathrm{w}}+\frac{1}{\mathrm{b}+\mathrm{w}}+\frac{1}{\mathrm{c}+\mathrm{w}}& =& \frac{2}{\mathrm{w}}\\ \frac{1}{\mathrm{a}+{\mathrm{w}}^2}+\frac{1}{\mathrm{b}+{\mathrm{w}}^2}+\frac{1}{\mathrm{c}+{\mathrm{w}}^2}& =& \frac{2}{{\mathrm{w}}^2}\end{array}$

$\Rightarrow \mathrm{w}, {\mathrm{w}}^2$  are roots of the equation  $\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{b}+\mathrm{x}}+\frac{1}{\mathrm{c}+\mathrm{x}}=\frac{2}{\mathrm{x}}$

$\begin{array}{rcl}\frac{\mathrm{a}+\mathrm{b}+2\mathrm{x}}{(\mathrm{a}+\mathrm{x}) (\mathrm{b}+\mathrm{x})}& =& \frac{2(\mathrm{c}+\mathrm{x})-\mathrm{x}}{\mathrm{x}(\mathrm{c}+\mathrm{x})}\end{array}$

$\begin{array}{rcl}\mathrm{x}(\mathrm{c}+\mathrm{x}) (\mathrm{a}+\mathrm{b}+2\mathrm{x})& =& (\mathrm{a}+\mathrm{x}) (\mathrm{b}+\mathrm{x}) (2\mathrm{c}+\mathrm{x})\end{array}$

$\begin{array}{rcl}\left(\mathrm{cx}+{\mathrm{x}}^2\right) \left(\mathrm{a}+\mathrm{b}+2\mathrm{x}\right)& =& 2\left({\mathrm{x}}^2+\mathrm{x}(\mathrm{a}+\mathrm{b})+\mathrm{ab}\right) \left(\mathrm{c}+\mathrm{x}\right)\end{array}$

$\begin{array}{rcl} & \Rightarrow & {\mathrm{x}}^3+(\mathrm{a}+\mathrm{b})\mathrm{cx}+\mathrm{abx}-2\mathrm{abc}=0\end{array}$

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

$\because \text{ Coefficient of }{x}^2=0\text{, the sum of roots }=0$

$\mathrm{w}+{\mathrm{w}}^2+\mathrm{\alpha }=0 \Rightarrow \mathrm{\alpha }-1=0$

$\therefore \mathrm{\alpha }=1$ is a root of $\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{b}+\mathrm{x}}+\frac{1}{\mathrm{c}+\mathrm{x}}=\frac{2}{\mathrm{x}}$

$\begin{array}{rcl} & \Rightarrow & \frac{1}{\mathrm{a}+1}+\frac{1}{\mathrm{b}+1}+\frac{1}{\mathrm{c}+1}=2\end{array}$