A) 2
B) 4
C) 6
D) 8
Correct Answer: A
Solution :
\[\therefore \]\[x+\frac{1}{x}=2\] Cubing both sides, \[{{\left( x+\frac{1}{x} \right)}^{3}}={{(2)}^{3}}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\times x\times \frac{1}{x}\left( x+\frac{1}{x} \right)=8\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\times 2=8\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=8-6=2\] Squaring both sides, \[{{\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)}^{2}}={{(2)}^{2}}\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}+2\times {{x}^{3}}\times \frac{1}{{{x}^{3}}}=4\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}+2=4\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}=4-2\] \[\Rightarrow \] \[{{x}^{6}}+\frac{1}{{{x}^{6}}}=2\]
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