A) 1340
B) 1364
C) 1358
D) 1360
Correct Answer: B
Solution :
(b): We know that \[{{\left( x-\frac{1}{x} \right)}^{2}}={{x}^{2}}+\frac{1}{{{x}^{2}}}-2\Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}=123-2\]\[\Rightarrow {{\left( x-\frac{1}{x} \right)}^{2}}=121\] \[{{\left( x-\frac{1}{x} \right)}^{2}}={{11}^{2}}\,\,\,\,\,\,\,\,\,\Rightarrow x-\frac{1}{x}=11\] \[\Rightarrow \left( x+\frac{1}{x} \right)={{11}^{3}}\] \[\Rightarrow {{x}^{3}}-\frac{1}{{{x}^{3}}}-3\left( x-\frac{1}{x} \right)=1331\] \[\Rightarrow {{x}^{3}}-\frac{1}{{{x}^{3}}}-3\times 11=1331\Rightarrow {{x}^{3}}-\frac{1}{{{x}^{3}}}=1331+33\Rightarrow {{x}^{3}}-\frac{1}{{{x}^{3}}}=1364\]
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