Question

if x=root3-2 find the value of x+1/xcube

Answer 1

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

Now,

${(x + \frac{1}{x}) }^{3} = {( \sqrt{3} - 2 + - 2 - \sqrt{3} )}^{2} \\ \\ = ( {0 -4)}^{3} \\ \\ = - 64$