If $$x=(9+4\sqrt{5})^{48} = [x] +f$$, where [x] is defined as integral part of x and f is a fraction, then x (1 - f) equals

It is given that $$x=(9+4\sqrt{5})^{48}$$ ... (1)

Let us assume that $$y=(9-4\sqrt{5})^{48}$$ ... (2)

We can see that  0 < y < 1. 

Also, x + y = 2($$48C0*(9)^{48}$$+$$48C2*(9)^{46}*(4\sqrt{5})^2$$+$$48C4(9)^{44}*(4\sqrt{5})^4$$+...+$$48C48*(4\sqrt{5})^{48}$$)

We can see that x + y is an integer therefore we can say that y + f = 1. Hence, y = (1 - f)

We can see that = x(1 - f) = x*y

$$\Rightarrow$$ $$(9+4\sqrt{5})^{48}(9-4\sqrt{5})^{48}$$

$$\Rightarrow$$ $$(9^2-(4\sqrt{5})^2)^{48}$$

$$\Rightarrow$$ $$(81-80)^{48}$$ = 1

Hence, option A is the correct answer.