Question

$if \: x - \frac{1}{x} = 8 \: find \: {x }^{2} + \frac{1}{ {x}^{2} }$
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Answer 1

Step-by-step explanation:

$x - \frac{1}{x} = 8$

Squaring both sides we get

${(x - \frac{1}{x} )}^{2} = {8}^{2}$

${x}^{2} + \frac{1}{ {x}^{2} } - 2x \times \frac{1}{x} = 64$

${x}^{2} + \frac{1}{ {x}^{2} } = 64 + 2 = 66$