If ‘f ’ is a polynomial such that $\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)\mathrm{f}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)=\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)+\mathrm{f}\left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)(\text{ where }\mathrm{x}\ne 0,\pm 1)\text{ and }\mathrm{f}\left(3\right)=28$ $\text{ then the value of }\frac{1}{605}\left(\sum _{\mathrm{n}=1}^{10} \left(\mathrm{f}\right(\mathrm{n})-1)\right)\text{ is }$

$\text{ replace }\frac{1-\mathrm{x}}{1+\mathrm{x}}\text{ ' by "x" we have }$
$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)\cdot \mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\pm {\mathrm{x}}^{\mathrm{n}}+1,\mathrm{f}\left(3\right)=28\\ \Rightarrow f\left(3\right)=3^{\mathrm{n}}+1=28\therefore n=3,f\left(x\right)=x^3+1\end{array}$
$\text{ Now }\sum _{\mathrm{n}=1}^{10} \left(\mathrm{f}\right(\mathrm{n})-1)=\sum _{\mathrm{n}=1}^{10} {\mathrm{n}}^3={\left(\frac{10\left(11\right)}{2}\right)}^2=3025$
$\Rightarrow \frac{1}{605}\left(3025\right)=5$