If $$\sqrt x - \frac{1}{\sqrt x} = \sqrt6$$, then $$x^2 + \frac{1}{x^2}$$ is equal to:
$$\sqrt x - \frac{1}{\sqrt x} = \sqrt6$$
Squaring both side,
$$\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2-2=\left(\sqrt{6}\right)^2$$
$$x+\frac{1}{x}-2=6$$
$$x+\frac{1}{x}=8$$
Again squaring both side,
$$\left(x+\frac{1}{x}\right)^2=8^2$$
$$\left(x^2+\frac{1}{x^2}+2=64\right)$$
$$\left(x^2+\frac{1}{x^2}=62\right)$$
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