Question : If $\left(x+\frac{1}{x}\right)=\sqrt{6}$ and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)$?
Option 1: $120\sqrt{3}$
Option 2: $128\sqrt{3}$
Option 3: $112\sqrt{3}$
Option 4: $108\sqrt{3}$
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Correct Answer: $112\sqrt{3}$
Solution : $\left(x+\frac{1}{x}\right)=\sqrt{6}$ Squaring both sides, $\left(x^2+\frac{1}{x^2}+2\right)=6$ ⇒ $x^4-4x^2+1=0$ ⇒ $x^2=\frac{4\pm2\sqrt{3}}2$ ⇒ $x^2 = 2+\sqrt{3}$ (Since $x>0$) Squaring both sides, ⇒ $x^4 = 7+4\sqrt{3}$ Squaring both sides ⇒ $x^8 = 97+56\sqrt{3}$ So, $\frac{1}{x^8}=97-56\sqrt{3}$ So, $x^8-\frac{1}{x^8}=112\sqrt{3}$ Hence, the correct answer is $112\sqrt{3}$.
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