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

$\mathrm{\omega },{\mathrm{\omega }}^2$  are roots of $\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{b}+\mathrm{x}}+\frac{1}{\mathrm{c}+\mathrm{x}}=\frac{2}{\mathrm{x}}$

$\begin{array}{l}\Rightarrow \frac{(b+x)(c+x)+(a+x)(c+x)+(a+x)(b+x)}{(a+x)(b+x)(c+x)}=\frac{2}{x}\\ \Rightarrow x\left[3x^2+2(a+b+c)x+(bc+ca+ab)\right]\\ =2\left[abc+(bc+ca+ab)x+(a+b+c)x^2+x^3\right]\\ \Rightarrow x^3-(bc+ca+ab)x-2abc=0\end{array}$

$$\begin{gathered} Sum of the roots =\alpha +\omega +{\omega }^2=0 \\ \Rightarrow \alpha =1 \end{gathered}$$

$\Rightarrow \frac{1}{\mathrm{a}+1}+\frac{1}{\mathrm{b}+1}+\frac{1}{\mathrm{c}+1}=\frac{2}{1}=2$