Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Option 1: $\frac{13}{4}$
Option 2: $\frac{13}{8}$
Option 3: $\frac {17}{4}$
Option 4: $\frac{17}{8}$
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Correct Answer: $\frac{13}{8}$
Solution :
Given: $2x-\frac{2}{x}=1$
Dividing both sides by 2, we get,
⇒ $x-\frac{1}{x} = \frac{1}{2}$
Cubing both sides, we get,
⇒ $(x-\frac{1}{x})^3=(\frac{1}{2})^3$
⇒ $x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})=\frac{1}{8}$
⇒ $x^3-\frac{1}{x^3}-3×\frac{1}{2}=\frac{1}{8}$
⇒ $x^3-\frac{1}{x^3}=\frac{1}{8}+\frac{3}{2}$
$\therefore x^3-\frac{1}{x^3}=\frac{13}{8}$
Hence, the correct answer is $\frac{13}{8}$.
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