Question

step wise answer please​

Answer 1

$\sqrt{2} < \sqrt{3} < \sqrt{5}$

Now,

$\sqrt{2} = {2}^{ \frac{1}{2} }$

so,

$\sqrt[4]{2} = {2}^{ \frac{1}{8} }$

$\sqrt[3]{2} = {2}^{ \frac{1}{6} }$

and 1/6>1/8