Question

$if \: x = 2 + \sqrt{3} \: \\ then \: find \: {(x + \frac{1}{x}) }^{3}$
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Answer 1

$\huge\rm\color{midnightblue} Solution$

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

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

rationalise the denominator

$\sf \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

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

$\sf \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}$

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

now,

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

$\sf x + \frac{1}{x} = 4$

Hence,

$\sf {(x + \frac{1}{x}) }^{3} = {4}^{3}$

$\fbox{\sf {(x + \frac{1}{x}) }^{3} = 64}$

Answer 2

$\rm\huge\color{palevioletred}Answer:-$

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