If the two roots of the equation $\left(\text{c - 1}\right){\left({\text{x}}^{\text{2}}\text{+x+1}\right)}^{\text{2}}\text{-}\left(\text{c+1}\right)\left({\text{x}}^{\text{4}}{\text{+x}}^{\text{2}}\text{+1}\right)\text{=0}$  are real and  distinct and  $\text{f}\left(\text{x}\right)\text{=}\frac{\text{1-x}}{\text{1+x}}$ then  $\text{f}\left(\text{f}\left(\text{x}\right)\right)\text{+f}\left(\text{f}\left(\frac{1}{\text{x}}\right)\right)$  is

Given   $\frac{c+1}{c-1}=\frac{{(x^2+x+1)}^2}{x^4+x^2+1}=\frac{{(x^2+x+1)}^2}{(x^2+x+1)(x^2-x+1)}=\frac{x^2+x+1}{x^2-x+1}$
$\Rightarrow c=\frac{x^2+1}{x}=x+\frac{1}{x}=\text{f}\left(\text{f}\left(x\right)\right)+\text{f}\left(\text{f}\left(\frac{1}{x}\right)\right)$
$\because [\text{f}\left(\text{f}\left(x\right)\right)=\frac{1-\text{f}\left(x\right)}{1+\text{f}\left(x\right)}=\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}=x]$