Question : If $\left(x-\frac{1}{x}\right)^2=12$, what is the value of $\left(x^2-\frac{1}{x^2}\right)$, given that $x>0$?
Option 1: $6 \sqrt{2}$
Option 2: $8 \sqrt{3}$
Option 3: $6 \sqrt{3}$
Option 4: $8 \sqrt{2}$
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Correct Answer: $8 \sqrt{3}$
Solution : Given, $\left(x-\frac{1}{x}\right)^2=12$ ⇒ $x^2+\frac{1}{x^2}=12+2$ ⇒ $x^2+\frac{1}{x^2}+2=12+2+2$ ⇒ $x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=12+2+2$ ⇒ $(x+\frac{1}{x})^2=16$ ⇒ $(x+\frac{1}{x})=4$ And, $\left(x-\frac{1}{x}\right)=2\sqrt3$ Now,$\left(x^2-\frac{1}{x^2}\right)$ = $(x-\frac{1}{x})(x+\frac{1}{x})$ = $4\times2\sqrt3$ = $8\sqrt3$ Hence, the correct answer is $8\sqrt3$.
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Question : If $\left(x+\frac{1}{x}\right)=\sqrt{6}$ and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)$?
Question : If ${\left(x-\frac{1}{x}\right)=\sqrt{6}}$ and $x > 1$, what is the value of ${\left(x^8-\frac{1}{x^8}\right)}$?
Question : If $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right)$?
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Question : If $\left(x+\frac{1}{x}\right)=5$, and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)?$
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