Question : If $\left(x + \frac{1}{x}\right) = 2 \sqrt{2}$, what is the value of $\left(x^6 + \frac{1}{x^6}\right)$?
Option 1: 198
Option 2: 216
Option 3: 180
Option 4: 234
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Correct Answer: 198
Solution :
Given: $\left(x + \frac{1}{x}\right) = 2 \sqrt{2}$
Cubing both sides, we get,
$(x + \frac{1}{x})^{3} = (2\sqrt{2})^{3}$
$⇒x^{3} + \frac{1}{x^3} + 3×x×\frac{1}{x}(x + \frac{1}{x}) = 16\sqrt{2}$
$⇒x^{3} + \frac{1}{x^3} + 3 \times 2\sqrt{2} = 16\sqrt{2}$
$⇒x^{3} + \frac{1}{x^3} = 10\sqrt{2}$
Now, squaring both sides, we get,
$⇒(x^{3} + \frac{1}{x^3})^{2} = (10\sqrt{2})^{2}$
$⇒x^{6} + \frac{1}{x^6} + 2×x^3×\frac{1}{x^3} = 200$
$\therefore x^{6} + \frac{1}{x^6} = 198$
Hence, the correct answer is 198.
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