Question : If $(x-\frac{1}{x})^2=\sqrt3$, then the value of $(x^6+\frac{1}{x^6})$ equals:
Option 1: 90
Option 2: 100
Option 3: 110
Option 4: 120
Correct Answer: 110
Solution :
Given: $(x-\frac{1}{x})^2=\sqrt3$
Squaring both sides, we get,
$⇒(x-\frac{1}{x})^2=(\sqrt3)^2$
$⇒x^2+\frac{1}{x^2}–2=3$
$⇒x^2+\frac{1}{x^2}=5$
Cubing on both sides, we get,
$⇒(x^2+\frac{1}{x^2})^3=5^3$
$⇒x^6+\frac{1}{x^6}+3(x^2+\frac{1}{x^2})=125$
$⇒x^6+\frac{1}{x^6}+3\times5=125$
$⇒x^6+\frac{1}{x^6}=125-15$
$\therefore x^6+\frac{1}{x^6}=110$
Hence, the correct answer is 110.
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