Solution

Given $$p={ \left( 8+3\sqrt { 7 } \right) }^{ n }$$
and $$p=\left[ p \right] +f$$ (where$$[.]=$$GIF)
Sol. : Let$$N={ \left( 8-3\sqrt { 7 } \right) }^{ n }$$
$$\Rightarrow$$ Now, $${ \left( a+b \right) }^{ n }+{ \left( a-b \right) }^{ n }=2\left[ { ^{ n }{ C } }_{ 0 }{ a }^{ n }+{ ^{ n }{ C } }_{ 2 }{ a }^{ n-2 }-{ b }^{ 2 }+...{ b }^{ n } \right] $$
$$\Rightarrow$$ Take $$a=8$$ & $$b=3\sqrt { 7 } $$
$$\therefore p+N=2\left[ { ^{ n }{ C } }_{ 0 }{ 8 }^{ n }+{ ^{ n }{ C } }_{ 2 }{ 8 }^{ n-2 }-{ (3\sqrt { 7 } ) }^{ 2 }+.... \right] $$
$$=2.K$$ (where $$k$$ &$$I$$)
Now, $$0$$\therefore p+N=[p]+f+N=2K$$
$$[P]$$ is always integer & $$p+N$$ is an integer
$$=f+N$$ must be an integer.
But, $$0<(f+N)<2$$ $$\rightarrow f+N=1$$
$$\Rightarrow p(1-f)=p.N={ \left( 8+3\sqrt { 7 } \right) }^{ n }{ \left( 8-3\sqrt { 7 } \right) }^{ n }$$
$$\Rightarrow{ \left( 8^2+(3\sqrt { 7 })^2 \right) }^{ n }=1^n=1.$$
Hence, the answer is $$1.$$