Question : If $\left(x+\frac{1}{x}\right)=5$, and $x>1$, what is the value of $\left(x^8-\frac{1}{x^8}\right)?$
Option 1: $60605 \sqrt{21}$
Option 2: $60615 \sqrt{21}$
Option 3: $60705 \sqrt{21}$
Option 4: $60725 \sqrt{21}$
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Correct Answer: $60605 \sqrt{21}$
Solution : Given: $\left(x+\frac{1}{x}\right)=5$--------------(i) Now, $\left(x^8-\frac{1}{x^8}\right)$ $=(x^4-\frac{1}{x^4})$$(x^4+\frac{1}{x^4})$ $=(x^2-\frac{1}{x^2})$$(x^2+\frac{1}{x^2})$$(x^4+\frac{1}{x^4})$ $= (x-\frac{1}{x})$$(x+\frac{1}{x})$$(x^2+\frac{1}{x^2})$$(x^4+\frac{1}{x^4})$----------(ii) Squaring equation (i), we get, $(x+\frac{1}{x})^2 = 5^2$ $⇒(x^2+\frac{1}{x^2}+2) = 25$ $⇒(x^2+\frac{1}{x^2}) = 23$-----------------(iii) Squaring equation (iii), we get, $(x^2+\frac{1}{x^2})^2 = 23^2$ $⇒(x^4+\frac{1}{x^4}+2) = 529$ $⇒(x^4+\frac{1}{x^4}) = 527$---------------(iv) Subtracting 2 on both sides of equation (iii), we get, $⇒(x^2+\frac{1}{x^2}-2) = 23-2$ $⇒(x-\frac{1}{x})^2 = 21$ $⇒(x-\frac{1}{x}) = \sqrt{21}$--------------(v) Substituting (i), (iii), (iv), and (v) in equation(ii), $\therefore(x^8-\frac{1}{x^8}) = \sqrt{21}×5×23×527= 60605\sqrt{21}$ Hence, the correct answer is $60605\sqrt{21}$.
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