A) 2
B) 6
C) 3
D) 8
Correct Answer: C
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, \[{{x}^{30}}+{{x}^{27}}+{{x}^{36}}+{{x}^{33}}-1\] \[=1+1+1+1-1=3\]
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