If $R=(7+4\sqrt{3}{)}^{2n}=I+f,$where $I\in N$ and $0<f<1$ , then $R(1-f)$ equal 

Let $G=(7-4\sqrt{3}{)}^{2n}$ . Then ,

$R+G=(7+4\sqrt{3}{)}^{2n}+(7-4\sqrt{3}{)}^{2n}$

$\Rightarrow R+G=2\left\{{ }^{2n}{C}_0{7}^{2n}{+}^{2n}{C}_2{7}^{2n-2}(4\sqrt{3}{)}^2+\dots \right\}$

$\Rightarrow R+G=$  An even integer 

$\Rightarrow I+f+G=$  An even integer 

$\Rightarrow f+G=$ An even integer  

$\Rightarrow f+G=1[\because 0\le f<1,0<G<1\therefore 0<f+G<2\Rightarrow f+G=1]$

$\Rightarrow G=1-f$

$\therefore R(1-f)=RG=(7+4\sqrt{3}{)}^{2n}(7-4\sqrt{3}{)}^{2n}=1$