Question : If $x\left(3-\frac{2}{x}\right)=\frac{3}{x}$, then the value of $x^3-\frac{1}{x^3}$ is equal to:
Option 1: $\frac{8}{27}$
Option 2: $\frac{61}{27}$
Option 3: $\frac{62}{27}$
Option 4: $\frac{52}{27}$
Correct Answer: $\frac{62}{27}$
Solution :
$x\left(3-\frac{2}{x}\right)=\frac{3}{x}$
⇒ $3x-2 = \frac{3}{x}$
⇒ $3x- \frac{3}{x} = 2$
⇒ $(x- \frac{1}{x}) = \frac{2}{3}$
$\because$ $(x-y)^3 = x^3-y^3-3xy(x-y)$
$(x- \frac{1}{x})^3 = (\frac{2}{3})^3$
⇒ $x^3-\frac{1}{x^3} - 3(x-\frac{1}{x}) = \frac{8}{27}$
⇒ $x^3-\frac{1}{x^3} - 3(\frac{2}{3}) = \frac{8}{27}$
⇒ $x^3-\frac{1}{x^3} = \frac{8}{27}+2$
⇒ $x^3-\frac{1}{x^3} = \frac{62}{27}$
Hence, the correct answer is $\frac{62}{27}$.
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