A) \[1\]
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
\[x+\frac{1}{x}=1\] \[{{x}^{2}}+1=x\] \[{{x}^{2}}-x+1=0\] Multiplying both side by \[x+1\] \[(x+1)({{x}^{2}}-x+1)=0\] \[{{x}^{3}}+1=0\] \[{{x}^{3}}=-1\] Now, \[{{x}^{6}}={{({{x}^{3}})}^{2}}={{(-1)}^{2}}=1\]
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