Question

If x = 2 + root3, find (x+1÷x)power 3
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Answer 1

Given:

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

To find:

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

Solution:

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ rationalising \: the \: denominator \\ \\ = > \frac{1}{2 + \sqrt{3} } \frac{ \times (2 - \sqrt{3} )}{ \times (2 - \sqrt{3}) } \\ \\ = > \frac{2 - \sqrt{3} }{ {2}^{2} - ( \sqrt{3} ) {}^{2} } \\ \\ = > \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3}$

Now,

$x + \frac{1}{ x} = 2 + \sqrt{3} + (2 - \sqrt{3} ) \\ \\ = > x + \frac{1}{x} = {4} \\ \\ \\ </em><em>Now</em><em>,</em><em> </em><em>\</em><em>\</em><em>{(x + \frac{1 }{x} )}^{3} = (4) {}^{3} \\ \\ = > (x + \frac{1}{x} ) {}^{3} ={ 64}$

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