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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