Question

if x=1+ root 2, then show that: (x-1/x)^3= 8​

Answer 1

$\quad \footnotesize {(x - \frac{1}{x})}^{3}$

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

$\footnotesize = { \big( \frac{ {(1 + \sqrt{2} )}^{2} - 1 }{1 + \sqrt{2} } \big)}^{3}$

$\footnotesize = { \big( \frac{1 + 2 + 2 \sqrt{2} - 1}{1 + \sqrt{2} } \big)}^{3}$

$\footnotesize = { \big( \frac{2 + 2 \sqrt{2} }{1 + \sqrt{2} } \big)}^{3}$

$\footnotesize = { \big( \frac{2 \cancel{(1 + \sqrt{2} )}}{ \cancel{(1 + \sqrt{2})} } \big)}^{3}$

$= {(2)}^{3}$

$= 8$

Hence, proved!