Question

If x=√3+√2, then the value of x3−1x3 is equal to​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

$\sf{ {a}^{3} - {b}^{3} = {(a - b)}^{3} \: + 3ab(a - b)\: }$

GIVEN

$\sf{ x = \sqrt{3} + \sqrt{2} \: }$

TO DETERMINE

$\displaystyle \sf{ {x}^{3} - \frac{1}{ {x}^{3} } \: }$

CALCULATION

Here

$\displaystyle \sf{ \frac{1}{ {x} } = \frac{1}{ \sqrt{3} + \sqrt{2} } \: }$

$\implies \: \displaystyle \sf{ \frac{1}{ {x} } = \frac{ \sqrt{3} - \sqrt{2} }{ (\sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2}) } \: }$

$\implies \: \displaystyle \sf{ \frac{1}{ {x} } = \frac{ \sqrt{3} - \sqrt{2} }{ {( \sqrt{3}) }^{2} - {( \sqrt{2} \: ) }^{2} } \: }$

$\implies \: \displaystyle \sf{ \frac{1}{ {x} } = \frac{ \sqrt{3} - \sqrt{2} }{3 - 2 } }$

$\implies \: \displaystyle \sf{ \frac{1}{ {x} } = \sqrt{3} - \sqrt{2} }$

$\therefore \: \: \displaystyle \sf{ x - \frac{1}{ {x} } = \sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2} }$

$\therefore \: \: \displaystyle \sf{ x - \frac{1}{ {x} } = 2 \sqrt{2} }$

Hence

$\displaystyle \sf{ {x}^{3} - \frac{1}{ {x}^{3} } \: }$

$= \displaystyle \sf{ { \bigg( x - \frac{1}{x} \: \bigg)}^{3} + 3 .x. \frac{1}{x} (x - \frac{1}{x} ) \: }$

$= \displaystyle \sf{ { \bigg( x - \frac{1}{x} \: \bigg)}^{3} + 3 (x - \frac{1}{x} ) \: }$

$= \displaystyle \sf{ { ( 2\sqrt{2} \: ) }^{3} + 3 \times 2 \sqrt{2} \: }$

$= \displaystyle \sf{ 16 \sqrt{2} + 6 \sqrt{2} }$

$= \displaystyle \sf{ 22 \sqrt{2} }$

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