A) 140
B) 120
C) 190
D) 160
Correct Answer: A
Solution :
[a]| \[{{x}^{4}}+\frac{1}{{{x}^{4}}}=727\] |
| Add 2 on both side, we get, |
| \[{{x}^{4}}+\frac{1}{{{x}^{4}}}+2=727+2\] |
| \[\Rightarrow \]\[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}=729\]\[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}=\sqrt{729}\] |
| \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}+27\]\[\Rightarrow \]\[{{\left( x-\frac{1}{x} \right)}^{2}}+2=27\] |
| \[\Rightarrow \]\[\,x-\frac{1}{x}=\sqrt{25}\]\[\Rightarrow \]\[x-\frac{1}{x}=5\] |
| \[\therefore \] \[{{\left( x-\frac{1}{x} \right)}^{3}}={{x}^{3}}-\frac{1}{{{x}^{3}}}-3\,\left( x-\frac{1}{x} \right)\] |
| \[{{(5)}^{3}}={{x}^{3}}-\frac{1}{{{x}^{3}}}-3\cdot (5)\] |
| \[\Rightarrow \] \[{{x}^{3}}-\frac{1}{{{x}^{3}}}=125+15=140\] |
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