Question : If $x^2-5\sqrt{5}x+1=0$ and $x > 0$, what is the value of $(x^3-\frac{1}{x^3})$?
Option 1: 1331
Option 2: 1364
Option 3: 1296
Option 4: 1244
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Correct Answer: 1364
Solution : Given: $x^2-5\sqrt{5}x+1=0$, and $x >0$ Dividing by $x$ on both sides, we get $x-5\sqrt{5}+\frac{1}{x}=0$ ⇒ $x+\frac{1}{x}=5\sqrt{5}$ Squaring both sides, we get ⇒ $(x+\frac{1}{x})^2=(5\sqrt{5})^2=125$ We know that $(x-\frac{1}{x})^2=(x+\frac{1}{x})^2-4$ ⇒ $(x-\frac{1}{x})^2=125-4=121$ ⇒ $(x-\frac{1}{x})=11$ Now, $(x-\frac{1}{x})^3=x^3-\frac{1}{x^3}-3×x×\frac{1}{x}(x-\frac{1}{x})$ ⇒ $11^3=x^3-\frac{1}{x^3}-3×11$ ⇒ $x^3-\frac{1}{x^3}=1331+33$ $\therefore x^3-\frac{1}{x^3}=1364$ Hence, the correct answer is 1364.
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