Question : If $x^2-\frac{1}{x^2}=4 \sqrt{2}$, what is the value of $x^4-\frac{1}{x^4}?$
Option 1: $16 \sqrt{2}$
Option 2: $8\sqrt{2}$
Option 3: $24 \sqrt{2}$
Option 4: $32 \sqrt{2}$
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Correct Answer: $24 \sqrt{2}$
Solution : $x^2-\frac{1}{x^2}=4 \sqrt{2}$ Squaring both sides, we get $x^4+\frac{1}{x^4}-2=32$ ⇒ $x^4+\frac{1}{x^4}=34$ Adding 2 on both sides, we get $x^4+\frac{1}{x^4}+2=36$ ⇒ ($x^2+\frac{1}{x^2})^2 =6^2$ ⇒ $x^2+\frac{1}{x^2}=6$ Now, $x^4-\frac{1}{x^4} = (x^2-\frac{1}{x^2} )(x^2+\frac{1}{x^2})$ $= 4 \sqrt{2}\times 6 =24\sqrt{2}$ Hence, the correct answer is $24\sqrt{2}$.
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