| If \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\frac{17}{4},\]then what is \[\left( {{x}^{3}}-\frac{1}{{{x}^{3}}} \right)\]equal to? |
A) \[\frac{75}{16}\]
B) \[\frac{63}{8}\]
C) \[\frac{95}{8}\]
D) None of these
Correct Answer: B
Solution :
| \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)=\frac{17}{4}\] |
| \[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2-2=\frac{17}{4}\] |
| \[\Rightarrow \]\[{{\left( x-\frac{1}{x} \right)}^{2}}+2=\frac{17}{4}\] |
| \[\Rightarrow \]\[{{\left( x-\frac{1}{x} \right)}^{2}}=\frac{17}{4}-2\]\[\Rightarrow \]\[\left( x-\frac{1}{x} \right)=\frac{3}{2}\] |
| On cubing both sides, we get |
| \[{{\left( x-\frac{1}{x} \right)}^{3}}={{\left( \frac{3}{2} \right)}^{3}}\] |
| \[\Rightarrow \]\[{{x}^{3}}-\frac{1}{{{x}^{3}}}-3\times \frac{1}{x}\cdot x\left( x-\frac{1}{x} \right)=\frac{27}{8}\] |
| \[\Rightarrow \] \[{{x}^{3}}-\frac{1}{{{x}^{3}}}=\frac{27}{8}+3\times \left( \frac{3}{2} \right)\] |
| \[\Rightarrow \]\[{{x}^{3}}-\frac{1}{{{x}^{3}}}=\frac{27}{8}+\frac{9}{2}\]\[\Rightarrow \]\[\left( {{x}^{3}}-\frac{1}{{{x}^{3}}} \right)=\frac{63}{8}\] |
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