A) \[{{p}^{\frac{3}{2}}}\]
B) \[\left( p+1 \right)\sqrt{p+2}\]
C) \[\left( p-1 \right)\sqrt{p+2}\]
D) \[\left( p+1 \right)\sqrt{p-2}\]
Correct Answer: C
Solution :
[c] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=P\] \[\Rightarrow \,\,{{\left( x+\frac{1}{1} \right)}^{2}}=p+2\] \[\therefore \,\,\,\left( x+\frac{x}{x} \right)={{(p+2)}^{\frac{1}{2}}}\] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3\left( x+\frac{1}{x} \right)\] \[={{(p+2)}^{\frac{3}{2}}}-3{{(p+2)}^{\frac{1}{2}}}\] \[=(p-1)\sqrt{p+2}\]
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