If $x^2 +\frac{1}{x^2}= 66$, the value of $x-\frac{1}{x}$ is _____.

10 8 9 6

8

then x2 + \(\frac{1}{x^2}\) = b Then, x - \(\frac{1}{x}\) = \(\sqrt {b - 2}\) If $x^2 +\frac{1}{x^2}= 66$ Then, x - \(\frac{1}{x}\) = \(\sqrt {66 - 2}\) = 8