Question

if X equals to 2 + root 3 find x plus one upon X ka whole cube​

Answer 1

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

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

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

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,

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

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

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

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