If $$x^8 - 1442 x^4 + 1 = 0$$, then possible value of $$x - \frac{1}{x}$$ is:
$$x^8 - 1442 x^4 + 1 = 0$$
$$x^4 - 1442 + \frac{1}{x^4} = 0$$
$$x^4 + \frac{1}{x^4} =Â 1442 $$
$$x^4 + \frac{1}{x^4} + 2Â = 1442 + 2$$
$$(x^2 + \frac{1}{x^2})^2 = (38)^2 $$
$$x^2 + \frac{1}{x^2} = 38 $$
$$x^2 + \frac{1}{x^2} - 2Â = 38 - 2 $$
$$(x - \frac{1}{x})^2 = 6^2 $$
$$x - \frac{1}{x} = 6$$
($$\because (a - b)^2 = a^2 - 2ab + b^2$$)
$$\therefore$$ The correct answer is option D.
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