A) 927
B) 414
C) 364
D) 322
Correct Answer: D
Solution :
\[\therefore \] \[x+\frac{1}{x}=3\,\,\,\Rightarrow \,\,{{\left( x+\frac{1}{x} \right)}^{3}}=27\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\times x\times \frac{1}{x}\,\left( x+\frac{1}{x} \right)=27\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=18\] \[\Rightarrow \] \[{{\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)}^{2}}=324\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}+2=324\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}=322\]
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