The correct option is A 1
Given p=(8+3√7)n
Assuming p1=(8−3√7)n
Clearly, 0<p1<1
Now,
p+p1=(8+3√7)n+(8−3√7)n=2[ nC08n+ nC28n−2(3√7)2+…]=2k, k∈Z
Given
p=[p]+f⇒[p]+f+p1=2k
⇒f+p1=2k−[p]⋯(1)
Which is integer
We know that,
p1∈(0,1), f∈(0,1)⇒0<p1+f<2
Using equation (1), the only interger possible
⇒f+p1=1⇒p1=1−f
Therefore,
p(1−f)=pp1=[(8+3√7)n(8−3√7)n]=(64−9×7)n=1