Question

If x-1/x =3+2√2,find the value of x^3 -1/x^3

Answer 1

ANSWER

$x - \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ take \: cube \: on \: both \: side \\ \\ {(x - \frac{1}{x}) }^{3} = {(3 - 2 \sqrt{2} )}^{3} \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 3(x)( \frac{1}{x} )(x - \frac{1}{x} ) = ({3 - 2 \sqrt{2}) }^{3} \\ \\ {x}^{3} - \frac{1}{ x^{3} } - 3(3 - 2 \sqrt{2} ) = {(3 - 2 \sqrt{2}) }^{3} \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 3(3 - 2 \sqrt{2} ) + ( {3 - 2 \sqrt{2}) }^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = (3 - 2 \sqrt{2} )(3 + {(3 - 2 \sqrt{2} )}^{2} ) \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = (3 - 2 \sqrt{2} )(3 + 9 + 8 - 12 \sqrt{2} ) \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = (3 - 2 \sqrt{2} )(20 - 12 \sqrt{2} ) \\ \\ {x}^{3} - \frac{1}</strong><strong>{</strong><strong>x^</strong><strong>3</strong><strong>} = 60 - 36 \sqrt{2} - 40 \sqrt{2} + 48 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 108 - 76 \sqrt{2}$

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