Irodyti ar ar reiskinys lygus 1

³√(1 + √2) * (sesto laipsnio)√ (3-2√2) = 1
kaip cia mazdaug irodyti?

[tex]\sqrt[3]{1+\sqrt{2}} \cdot \sqrt[6]{3-2\sqrt{2}}= 1[/tex]
[tex]\left ( 1+\sqrt{2} \right )\left ( \sqrt{3-2\sqrt{2}} \right ) = 1[/tex]
[tex]\left ( 3+2\sqrt{2} \right )\left ( 3-2\sqrt{2} \right ) = 1[/tex]
[tex]9-6\sqrt{2}+6\sqrt{2}-8= 1[/tex]
[tex]1 = 1[/tex]