Let the function f (x) is given by f(x)=a0xn+a1xn−1+a2xn−2+.....+an−1x+an The given relation is f(x).(
1
x
)=f(x)+f(
1
x
) Substitute f(x) in the given equation (a0xn+a1xn−1+....+an)(a0xn+a1xn−1+....+an) Compare the coefficient of xn, a0an=a0⇒an=1 Compare the coefficient of xn−1 a0an−1+ana1=a1 a0an−1=0 So, an−1=an−2=.....=a1=0 a0=±1 f(x)=1±xn f(4)=1±4n=257 4n=256 n=4 Therefore, f(3)=1+34=82