Question

If x=3+2✓2,Then prove that (✓x)³-(1/✓x)³=14

Answer 1

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

LHS =  $\sqrt{x} ^{3} - \frac{1}{\sqrt{x} ^{3} }$

       Let $\sqrt{x} = p$

       $x = p^{2}$

     

Therefore,

 $\sqrt{x} ^{3} - \frac{1}{\sqrt{x} ^{3} }$

= $p^{3} - \frac{1}{p^{3} }$

= $(p - \frac{1}{p})(p^{2} + \frac{1}{p^{2} } + p*\frac{1}{p} )$

= $\frac{p^{2}-1 }{p} (p^{2} +\frac{1}{p^{2} } +1)$

Substituting    $x = p^{2}$

⇒$\frac{x-1}{\sqrt{x} } (x+\frac{1}{x}+1)$

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

⇒$\frac{3+\sqrt{2}-1}{\sqrt{3+\sqrt{2} }} (\frac{(3+\sqrt{2}) ^{2}+1+3+\sqrt{2} }{3+\sqrt{2} })$

= $\frac{2+\sqrt{2} }{\sqrt{3+\sqrt{2} } }(\frac{9+2+6\sqrt{2}+4+\sqrt{2} }{3+\sqrt{2} })$

= $\frac{2+\sqrt{2} }{\sqrt{3+\sqrt{2} } }(\frac{15+7\sqrt{2}}{3+\sqrt{2} })$

= $\frac{30+15\sqrt{2}+14\sqrt{2}+28 }{\sqrt{3+\sqrt{2} } }*\frac{1}{3+\sqrt{2} }$

= $\frac{58+29\sqrt{2} }{\sqrt{3+\sqrt{2} } } *\frac{1}{3+\sqrt{2} }$

= $\frac{29(2+\sqrt{2}) }{(3+\sqrt{2}) ^{\frac{3}{2} } }$

Please verify the question again

HOPE IT HELPS !!