If $$x + \frac{1}{x} = 8$$,then $$x^2 + \frac{1}{x^2}$$ is equalto:
$$\left(x+\frac{1}{x}\right)^{2\ }=x^2+\frac{1}{x^2}+2x.\ \frac{1}{x}$$
$$64=x^2+\frac{1}{x^2}+2$$
$$64\ -\ 2=x^2+\frac{1}{x^2}$$
$$62=x^2+\frac{1}{x^2}$$
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