A) 122
B) 118
C) 123
D) 120
Correct Answer: C
Solution :
\[x+\frac{1}{x}=3\] On squaring, \[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=9\] \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=7\] Again, \[{{\left( x+\frac{1}{x} \right)}^{3}}=27\] \[\Rightarrow \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)=27\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+9=27\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=18\] \[\therefore \] \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)\,\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)=7\times 18\] \[\Rightarrow \] \[{{x}^{5}}+x+\frac{1}{x}+\frac{1}{{{x}^{5}}}=126\] \[\Rightarrow \] \[{{x}^{5}}+\frac{1}{{{x}^{5}}}+3=126\] \[\Rightarrow \] \[{{x}^{5}}+\frac{1}{{{x}^{5}}}=123\]
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