If f (x) is a polynomial satisfying  $f\left(x\right)\cdot f(1/x)=f\left(x\right)+f(1/x)$  and $f\left(3\right)=28,$ then $f\left(4\right)$ s given by 

By considering a general nth degree polynomial and writing the expression

$f\left(x\right)\cdot f(1/x)=f\left(x\right)+f(1/x)$ in  terms of it, it can be proved by comparing te coefficients of  $x^n,x^{n-1},\cdots$

and the constant term, that the polynomial satisfying the above equation is either of the form  $x{ }^n+1$ 

or   $-x^n+1$ Now, from $f\left(3\right)=3^n+1=28$ , we get $3^n=27$  or $n=3.$ But  $f\left(3\right)=-3^n+1=28$ 

is not possible, as  $-3^n=$ 27 is not true for any value of n. Hence $f\left(4\right)=4^3+1=65$