Question : If $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right)$?
Option 1: $140\sqrt{2}$
Option 2: $116\sqrt{2}$
Option 3: $144\sqrt{2}$
Option 4: $128\sqrt{2}$
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Correct Answer: $140\sqrt{2}$
Solution : $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ Squaring both sides. ⇒ $ (x+\frac{1}{x})^2= 8$ ⇒ $(x-\frac{1}{x})^2 = (x+\frac{1}{x})^2-4$ ⇒ $(x-\frac{1}{x})^2=8-4=4$ ⇒ $(x-\frac{1}{x})= 2$ So, $(x^2-\frac{1}{x^2})=(x+\frac{1}{x})(x-\frac{1}{x})= 2\sqrt{2}×2=4\sqrt{2}$ Cubing both sides, we get ⇒ $x^6-\frac{1}{x^6}-3(x^2-\frac{1}{x^2})=64×(2\sqrt{2})$ ⇒ $x^6-\frac{1}{x^6}-3(4\sqrt{2})=128\sqrt{2}$ ⇒ $x^6-\frac{1}{x^6}=140\sqrt{2}$ Hence, the correct answer is $140\sqrt{2}$.
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