A) \[118\frac{1}{2}\]
B) \[30\frac{10}{27}\]
C) 0
D) 1
Correct Answer: B
Solution :
\[3x+\frac{1}{2x}=5\] On multiplying both sides by\[\frac{2}{3},2x+\frac{1}{3x}=\frac{10}{3}\] Cubing both sides, \[8{{x}^{2}}+\frac{1}{27{{x}^{3}}}+3\times 2x\times \frac{1}{3x}\] \[\left( 2x+\frac{1}{3x} \right)=\frac{1000}{27}\] \[\Rightarrow \] \[8{{x}^{3}}+\frac{1}{27{{x}^{3}}}+2\times \frac{10}{3}=\frac{1000}{27}\] \[\Rightarrow \] \[8{{x}^{3}}+\frac{1}{27{{x}^{3}}}=\frac{1000}{27}-\frac{20}{3}\] \[\therefore \] \[8{{x}^{3}}+\frac{1}{27{{x}^{3}}}=\frac{1000}{27}-\frac{20}{3}\] \[=\frac{1000-180}{27}=\frac{820}{27}=30\frac{10}{27}\]You need to login to perform this action.
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