If $\mathrm{p}=(8+3\sqrt{7}{)}^{\mathrm{n}}$ and $\mathrm{f}=\mathrm{p}-\left[\mathrm{p}\right]$, where [.] denotes the greatest integer function, then the value of p(1 - f) is equal to

$\begin{array}{l}\mathrm{p}=(8+3\sqrt{7}{)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{8}^{\mathrm{n}}{+}^{\mathrm{n}}{\mathrm{C}}_1{8}^{\mathrm{n}-1}(3\sqrt{7})+\dots \\ 49<63<64\Rightarrow 7<\sqrt{63} <8 \Rightarrow -8<-3\sqrt{7} < -7\Rightarrow 0 <8-3\sqrt{7}< 1\\ 0^n<{(8-3\sqrt{7})}^n<1^n \Rightarrow 0<{(8-3\sqrt{7})}^n<1\\ \text{ Let }{\mathrm{p}}_1=(8-3\sqrt{7}{)}^{\mathrm{n}} \\ {\mathrm{p}}_1=(8-3\sqrt{7}{)}^{\mathrm{n}}{=}^{\mathrm{n}}{\mathrm{C}}_0{8}^{\mathrm{n}}{-}^n{\mathrm{C}}_1{8}^{\mathrm{n}-1}(3\sqrt{7})+\dots \\ \mathrm{p}+{\mathrm{p}}_1=2\left({ }^{\mathrm{n}}{\mathrm{C}}_0{8}^{\mathrm{n}}{+}^{\mathrm{n}}{\mathrm{C}}_2{8}^{\mathrm{n}-2}(3\sqrt{7}{)}^2+\cdots \right)=\text{ even integer }\\ {\mathrm{p}}_1\text{ clearly belongs to }(0,1) \\ \Rightarrow \left[\mathrm{p}\right]+\mathrm{f}+{\mathrm{p}}_1=\text{ even integer }\\ \Rightarrow \mathrm{f}+{\mathrm{p}}_1=\text{ integer }\\ \mathrm{f}\in (0,1),{\mathrm{p}}_1\in (0,1)\\ \Rightarrow \mathrm{f}+\mathrm{p}\in (0,2)\\ \Rightarrow \mathrm{f}+{\mathrm{p}}_1=1\\ \Rightarrow {\mathrm{p}}_1=1-\mathrm{f}\end{array}$

Now $\mathrm{p}(1-\mathrm{f})={\mathrm{pp}}_1={\left[(8+3\sqrt{7}{)}^{\mathrm{n}}(8-3\sqrt{7})\right]}^{\mathrm{n}}=1$