Question

If x=given then find the give x^3

Answer 1

$x = \sqrt[3]{2 + \sqrt{3} } \\ = > {x}^{3} = 2 + \sqrt{3} - - - - - - 1 \\ \\ = > \frac{1}{x} = \frac{1}{ \sqrt[3]{2 + \sqrt{3} } } \\ = > \frac{1}{ {x}^{3} } = \frac{1}{2 + \sqrt{3} } \\ = > \frac{1}{ {x}^{3} } = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ = > \frac{1}{ {x}^{3} } = \frac{2 - \sqrt{3} }{4 - 3} = \frac{2 - \sqrt{3} }{1} - - - - - 2 \\ \\ \\ from \: \: eq.1 \: \: and \: \: 2 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } =2 + \sqrt{3} + 2 - \sqrt{3} = 4.$



Therefore answer is 4.



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