A) 125
B) 140
C) 155
D) 170
Correct Answer: B
Solution :
\[\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)=727\] \[\because \,\,\,\,\,{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+\frac{1}{{{x}^{4}}}+2\] \[=727+2=729\,\,\,\,\Rightarrow \,\,\,\,{{x}^{2}}+\frac{1}{{{x}^{2}}}=27\] \[{{\left( x-\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}-2\] \[\therefore \,\,\,\,\,\,{{\left( x-\frac{1}{x} \right)}^{2}}=27-2\Rightarrow x-\frac{1}{x}=5\] \[{{\left( x-\frac{1}{x} \right)}^{3}}={{x}^{3}}-\frac{1}{{{x}^{3}}}-3\left( x-\frac{1}{x} \right)\] \[{{(5)}^{3}}={{x}^{3}}-\frac{1}{{{x}^{3}}}-3\times 5\] \[125+15={{x}^{3}}-\frac{1}{{{x}^{3}}}\] \[\therefore \,\,\,{{x}^{3}}-\frac{1}{{{x}^{3}}}=\underline{\mathbf{140}}\]
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