Question

Find [x-1/x]whole cubed where x=3+2under root 2

Answer 1

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

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

$\rm{rationalise \: the \: denominator}$

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

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

$\star \rm{note : \:{(2 \sqrt{2})}^{2} = {2}^{2} \times {( \sqrt{2} )}^{2} } \\ = {4 \times 2 = 8} \\ \implies\rm{ \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ 9 - 8 }}$

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

$\rm{now}$

$\rm{x - \frac{1}{x} =3 + 2 \sqrt{2} - (3 - 2 \sqrt{2} )}$

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

$\implies\rm{x - \frac{1}{x} =4 \sqrt{2} } \\\implies \rm{ {(x - \frac{1}{x}) }^{3} = {(4 \sqrt{2} )}^{3} }$

$\fbox{\rm{ {( x - \frac{1}{x}) }^{3} =128\sqrt{2} }}$