Question : If $\left(x - \frac{1}{x}\right) = 2 \sqrt{2}$, what is the value of $\left(\mathrm{x}^6 + \frac{1}{\mathrm{x}^6}\right)$?
Option 1: 1030
Option 2: 960
Option 3: 1000
Option 4: 970
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Correct Answer: 970
Solution :
Given: $\left(x - \frac{1}{x}\right) = 2 \sqrt{2}$
We know, $(x - \frac{1}{x})^{3} = x^{3} - \frac{1}{x^3} - 3×x×\frac{1}{x}(x - \frac{1}{x})$
$⇒(x - \frac{1}{x})^{3} = x^{3} - \frac{1}{x^3} - 3 \times 2\sqrt{2}$
$⇒(2\sqrt{2})^{3} = x^{3} - \frac{1}{x^3} -6\sqrt{2}$
$⇒x^{3} - \frac{1}{x^3} = 22\sqrt{2}$
$⇒(x^{3} - \frac{1}{x^3})^{2} = (22\sqrt{2})^{2}$
$⇒ x^{6} + \frac{1}{x^6} - 2×x^{3} × \frac{1}{x^3}= 484×2$
$⇒x^{6} + \frac{1}{x^6} = 968+2$
$\therefore x^{6} + \frac{1}{x^6} = 970$
Hence, the correct answer is 970.
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