Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Option 1: $20 \sqrt{2}$
Option 2: $24 \sqrt{2}$
Option 3: $18 \sqrt{2}$
Option 4: $22 \sqrt{2}$
Correct Answer: $22 \sqrt{2}$
Solution :
Given: $x^2 -\frac{1}{x^2} = 4\sqrt6$
Squaring both sides we get,
⇒ $x^4 +\frac{1}{x^4} - 2 = 96$
⇒ $x^4 +\frac{1}{x^4}+2 = 98+2$
⇒ $ (x^2+\frac{1}{x^2})^2 =100$
⇒ $x^2 +\frac{1}{x^2} = 10$ ---------------(1)
⇒ $x^2 +\frac{1}{x^2} -2= 10-2$
⇒ $(x-\frac{1}{x})^2=8$
⇒ $(x-\frac{1}{x}) = 2\sqrt2$ ----------------(2)
So, $x^3 -\frac{1}{x^3} = (x-\frac{1}{x})(x^2 +\frac{1}{x^2}+ x×\frac{1}{x})$
⇒ $x^3 -\frac{1}{x^3}= 2\sqrt2 × (10+1)= 22\sqrt2$
Hence, the correct answer is $22\sqrt2$.
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