Question

x=
$\sqrt[3]{2 + \sqrt{3} }$
then
$x {3}^{} + \binom{1}{x} 3 =$

Answer 1

$\textbf{ Hello Mate! }$

$x = \sqrt[3]{2 + \sqrt{3} } \\ {x}^{3} = { (\sqrt[3]{2 + \sqrt{3} } )}^{3} \\ {x}^{3} = 2 + \sqrt{3} \\ \frac{1}{ {x}^{3} } = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ = \frac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3} \\ {x}^{3} + \frac{1}{{x}^{3}} = 2 + \sqrt{3 } + 2 - \sqrt{3} \\ = 4$

$\boxed{ Answer\:is\: 4 }$

Have great future ahead!
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