If $x = (7 + 4\sqrt {3})^{2n} = [x] + f$, where $n \epsilon N$ and $0\leq f < 1$, then $x(1 - f)$ is equal to

$\begin{array}{l} x=(7+4 \sqrt{3})^{2 n} \\ x=3+f \end{array}$

$f$ is fractional part of $x$

$\Rightarrow 0<f<1$

Let,

$f^{\prime} =(7-4 \sqrt{3})^{2 n} \\ 0 <7-4 \sqrt{3}<1 \\ 0 <(7-4 \sqrt{3})^{n}<1 \\ 0 <f^{\prime} <1$

$\begin{aligned} 1+f+f^{\prime} & =2\left[{ }^{2 n} c_{0}(7)^{2 n}+2 n c_{2} \cdot(7)^{2 n-2}+\cdots\right] \\ & =\text { even } \end{aligned}$

$0+f+f'<2$ is integer.

'$I$ ' is integer and $f+f'$ ' is also integer

$\begin{array}{c}\Rightarrow f+f^{\prime}=1 \\ f^{\prime}=1-f\end{array}$

$x(1-f)=x f^{\prime}=(7+4 \sqrt{3})^{2 n}(7-4 \sqrt{3})^{2 n}=1$