Discussion
If \(x^{4} + \frac{1}{x^{4}} = 322\), \(x \neq 0\), then one of the values of \(\left(x - \frac{1}{x}\right)\) is
Answer: B
One value of \(\left(x - \frac{1}{x}\right)\) is 4.
Use the identity \(\left(t - \frac{1}{t}\right)^{2} = t^{2} + \frac{1}{t^{2}} - 2\) and its plus form twice.
\[ \begin{aligned} \left(x^{2} + \tfrac{1}{x^{2}}\right)^{2} &= 322 + 2 = 324 \\ x^{2} + \tfrac{1}{x^{2}} &= 18 \\ \left(x - \tfrac{1}{x}\right)^{2} &= 18 - 2 = 16 \end{aligned} \]
So \(x - \frac{1}{x} = \pm 4\), and 4 is one of its values. (\(x^{2} + \frac{1}{x^{2}}\) is positive, so it is 18, not \(-18\).)
Check: if \(x - \frac{1}{x} = 4\), then \(x^{2} + \frac{1}{x^{2}} = 16 + 2 = 18\) and \(x^{4} + \frac{1}{x^{4}} = 324 - 2 = 322\). The values 2, 6 and 8 give squares of 4, 36 and 64, none of which equals 16.
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