A) 4
B) 5
C) 8
D) 15
Correct Answer: B
Solution :
[b] \[2p+\frac{1}{p}=4\] \[\Rightarrow \,\,\,p+\frac{1}{2p}=2\] \[\therefore \,\,{{\left( p+\frac{1}{2p} \right)}^{3}}\] \[={{p}^{3}}+\frac{1}{8{{p}^{3}}}+3.p.\frac{1}{2p}\,\,\left( p+\frac{1}{2p} \right)\] \[\Rightarrow \,\,\,8={{p}^{3}}+\frac{1}{8{{p}^{3}}}+\frac{3}{2}\times 2\] \[\Rightarrow {{p}^{3}}+\frac{1}{8{{p}^{3}}}=8-\,3=5\]
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