A) \[-\frac{9}{8}\]
B) \[-\frac{25}{8}\]
C) \[\frac{7}{8}\]
D) \[11\]
Correct Answer: C
Solution :
| [c] \[2x+\frac{2}{x}=3\]\[\Rightarrow \]\[x+\frac{1}{x}=\frac{3}{2}\] \[\Rightarrow \] \[{{\left( x+\frac{1}{x} \right)}^{3}}={{\left( \frac{3}{2} \right)}^{2}}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)=\frac{27}{8}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3.\frac{3}{2}=\frac{27}{8}{{x}^{3}}+\frac{1}{{{x}^{3}}}=\frac{27}{8}-\frac{9}{2}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=-\frac{9}{8}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+2=2-\frac{9}{8}\] \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+2=\frac{7}{8}\] |
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