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$If \: x - \frac{1}{x} = \frac{1}{6} \: \: \: \: then, {x}^{3} + \frac{1}{ {x}^{3} }$



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Answer 1

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Part 1:

${x}^{2} + \dfrac{1}{ {x}^{2} } = {(x - \dfrac{1}{x}) }^{2} + 2 \\ \\ {x}^{2} + \dfrac{1}{ {x}^{2} } = {( \dfrac{1}{6} )}^{2} + 2 \\ \\ {x}^{2} + \frac{1 }{ {x}^{2} } = \dfrac{1}{36} + 2 = > \dfrac{1 + 72}{36} = > \dfrac{73}{36}$

Part 2:

${(x + \dfrac{1}{x}) }^{2} - 2 = {x}^{2} + \dfrac{1}{ {x}^{2} } \\ \\ {(x + \dfrac{1}{x}) }^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \\ \\ {(x + \dfrac{1}{x}) }^{2} = \dfrac{73}{36} + 2 = > \dfrac{73 + 72}{36} = > \frac{145}{36} \\ \\ (x + \dfrac{1}{x} ) = \sqrt{ \dfrac{145}{36} } = > \dfrac{ \sqrt{145} }{6}$

Now we know that,

$( {x}^{3} + \dfrac{1}{ {x}^{3} } ) = (x + \dfrac{1}{x} )( {x}^{2} + \dfrac{1}{ {x}^{2} } - 1) \\ \\ ( {x}^{3} + \dfrac{1}{ {x}^{3} } ) = \dfrac{ \sqrt{145} }{6}( \dfrac{73}{36} - 1) \\ \\ ( {x}^{3} + \frac{1}{ {x}^{3} } ) = \dfrac{ \sqrt{145} }{6} \times \dfrac{37}{36} = > \frac{37 \sqrt{145} }{216}$

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