Question

plz solve this question it is very important
$if \: x - \frac{1}{x } = 4 \: then \: find \: {x}^{3} - \frac{1}{ {x}^{3} }$
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Answer 1

Answer:

$x - \frac{1}{x} = 4 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } \\ = (x - \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } + x \times \frac{1}{x} ) \\ = 4 \times ( {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} - 2 \times x \times \frac{1}{x} + x \times \frac{1}{x} ) \\ = 4 \times (( {x - \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} + x \times \frac{1}{x} ) \\ = 4 \times ( {4}^{2} + 2 + 1) \\ = 4 \times (16 + 2 + 1) \\ = 4 \times 19 \\ = 76$