Answer:
Option B
Explanation:
Given surds are of order 2 and 3, Their LCM is 6.
Changing each to a surd of order 6, we get :
$\sqrt{2} = 2^{\frac{1}{2}}$
$=2^{\left(\frac{1}{2} \times \frac{3}{3}\right)}$
$=2^{\frac{3}{6}}$
$=(2^{3})^{\frac{1}{6}}$
$=(8)^{\frac{1}{6}}$
$=\sqrt[6]{8}$
$\sqrt[3]{3} = 3^{\frac{1}{3}}$
$=3^{\left(\frac{1}{3} \times \frac{2}{2}\right)}$
$=3^{\frac{2}{6}}$
$=(3^{2})^{\frac{1}{6}}$
$=(9)^{\frac{1}{6}}$
$=\sqrt[6]{9}$
Clearly, $\sqrt[6]{9} > \sqrt[6]{8}$ and hence $\sqrt[3]{3}> \sqrt{2}$
