Question : If $x^2-\sqrt{9.76} x+1=0$ and $x>1$, then the value of $\left(x^3-\frac{1}{x^3}\right)$ is:
Option 1: 21.042
Option 2: 24.024
Option 3: 21.024
Option 4: 24.042
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Correct Answer: 21.024
Solution :
Given, $x^2-\sqrt{9.76} x+1=0$
Dividing by $x$, we get,
$⇒x-\sqrt{9.76} +\frac{1}{x}=0$
$⇒x +\frac{1}{x}=\sqrt{9.76}$
Squaring both sides, we get,
$⇒x^2+\frac{1}{x^2}+2×x×\frac{1}{x}=9.76$
Subtracting 4 from both sides, we get,
$⇒x^2+\frac{1}{x^2}-2=9.76-4$
$⇒(x -\frac{1}{x})^2=5.76$
$⇒(x -\frac{1}{x})=\sqrt{5.76}$
Cubing both sides, we get,
$⇒ x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})=5.76×\sqrt{5.76}$
$⇒ x^3-\frac{1}{x^3}-3×\sqrt{5.76}=5.76×\sqrt{5.76}$
$⇒ x^3-\frac{1}{x^3}=5.76×\sqrt{5.76}+3×\sqrt{5.76}$
$⇒ x^3-\frac{1}{x^3}=8.76×\sqrt{\frac{576}{100}}=8.76×\frac{24}{10}=21.024$
Hence, the correct answer is 21.024.
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