Question

(vi) $\sqrt[8]{12},\sqrt[4]{6}$

Answer 1

The orders of $\sqrt[8]{12}\mathrm{and}\sqrt[4]{6}$ are 8 and 4.
The LCM of 8 and 4 is 8.
So, we will make the orders of both the surds 8.
$\sqrt[8]{12}\mathrm{and}\sqrt[4]{6}=\sqrt[8]{6^2}=\sqrt[8]{36}$
Hence, the order is same.
Since 12 < 36, $\sqrt[8]{12}<\sqrt[4]{6}$.