Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Option 1: $\frac{2 \sqrt{6}-6}{5}$
Option 2: $\frac{4 \sqrt{6}-6}{5}$
Option 3: $\frac{4 \sqrt{6}-6}{3}$
Option 4: $\frac{4 \sqrt{3}-6}{5}$
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Correct Answer: $\frac{4 \sqrt{6}-6}{5}$
Solution : Given: $x=(\sqrt{6}-1)^{\frac{1}{3}}$ Cubing both sides, we get, $x^3=\sqrt6-1$ $\therefore \frac{1}{x^3}=\frac{1}{\sqrt6-1}=\frac{\sqrt6+1}{5}$ Now, $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ $=x^3-\frac{1}{x^3}-3×x×\frac{1}{x^3}(x-\frac{1}{x})-3(x-\frac{1}{x})$ $=x^3-\frac{1}{x^3}$ Putting values, we get, $=(\sqrt6-1)-\frac{\sqrt6+1}{5}$ $=\frac{4 \sqrt{6}-6}{5}$ Hence, the correct answer is $\frac{4 \sqrt{6}-6}{5}$.
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