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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