If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ Problem : 
If $f(x) $ be a polynomial function satisfying $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$ and $f(4) =65$ then find $f(6)$ 
Solution : 
$f(x) f(\frac{1}{x})-f(x) =f(\frac{1}{x})$
$\Rightarrow f(x) =\frac{f(1/x)}{f(1/x)-1}$.....(i)
Also $f(x).f(\frac{1}{x})=f(x) +f(\frac{1}{x})$
$\Rightarrow f(\frac{1}{x})=\frac{f(x)}{f(x)-1}$ ......(ii)
On multiplying (i) and (ii) , we get 
$f(x) .f(\frac{1}{x})=\frac{f(1/x).f(x)}{(f(1/x)-1) ((f)(x)-1)}$
$\Rightarrow (f(\frac{1}{x}) -1)(f(x)-1)=1$ 
Please suggest how to proceed here since $f(x)-1 ; \& f(\frac{1}{x}-1)$ are reciprocal to each other 
Thanks

The equations are satisfied by $f(x) = x^3 + 1$.  For:
$$
    f(x) f\left(\frac{1}{x}\right)
    = (x^3  + 1) \left(\frac{1}{x^3} + 1\right)
    = 1 + x^3 + \frac{1}{x^3} + 1
    = f(x) + f \left(\frac{1}{x}\right)  
$$
and $f(4) = 4^3 + 1 = 65$.
So
$f(6) = 6^3 + 1 = 217$.
