Question : If $(x+\frac{1}{x})=6$ and $x>1$, find the value of $(x^2–\frac{1}{x^2})$.
Option 1: $18 \sqrt{2}$
Option 2: $30 \sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $12 \sqrt{10}$
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Correct Answer: $24 \sqrt{2}$
Solution : Given: $(x+\frac{1}{x})=6$, and $x>1$. We know, $(x–\frac{1}{x})^2=(x+\frac{1}{x})^2–4×x×\frac{1}{x}$ ⇒ $(x–\frac{1}{x})^2=(6)^2–4$ ⇒ $(x–\frac{1}{x})=\sqrt{32}=4\sqrt2$ ⇒ $x^2–\frac{1}{x^2}=(x+\frac{1}{x})(x–\frac{1}{x})$ ⇒ $x^2–\frac{1}{x^2}=(6)(4\sqrt2)=24\sqrt2$ Hence, the correct answer is $24\sqrt2$.
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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 $\left(x^2+\frac{1}{x^2}\right)=6$ and $0<x<1$, what is the value of $x^4-\frac{1}{x^4}$?
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 $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
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 $x>0$?
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