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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