If $x=(9+4\sqrt{5})^{48}=[x]+f$ . Find $x(1-f)$. 
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 .
$\color{green}{a.)\ 1} \\
b.)\ \text{less than}\ 1 \\
c.)\ \text{more than}\ 1 \\
d.)\ \text{between}\ 1 \text{and }\ 2 \\
e.)\ \text{none of these}\ \\ $

This question looks scary from the get go.
I tried to go with pattern
$(9+4\sqrt{5})^{1}\approx 17.94=17+0.94 \implies 17(1-0.94)=1.02\\
(9+4\sqrt{5})^{2}\approx 321.99=321+0.99 \implies 321(1-0.99)=3.21\\ $
i don't know if i interpreted the question correctly .
I look for a short and simple way.
I have studied maths upto $12$th grade. Thanks.

Consider
$$P=(9+4\sqrt 5)^{48}+(9-4\sqrt 5)^{48}.$$
Note that $P$ is an integer.
Now we have $0\lt 9-4\sqrt 5\lt 1$. Hence we have
$$0\lt (9-4\sqrt 5)^{48}\lt 1.$$
Hence, we have
$$x=(9+4\sqrt 5)^{48}=P-1+1-(9-4\sqrt 5)^{48}.$$
This implies that $f=1-(9-4\sqrt 5)^{48}$.
Thus, we have
$$x(1-f)=(9+4\sqrt 5)^{48}(9-4\sqrt 5)^{48}=1.$$
