Write in descending order :

(i) $2\sqrt[4]{6} \text{ and } 3\sqrt[4]{2}$

(ii) $7\sqrt{3} \text{ and } 3\sqrt{7}$

(i) Given,

$\Rightarrow 2\sqrt[4]{6} = \sqrt[4]{2^4 \times 6} = \sqrt[4]{96} \\[1em] \Rightarrow 3\sqrt[4]{2} = \sqrt[4]{3^4 \times 2} = \sqrt[4]{162}.$

Since, 162 > 96

$\therefore \sqrt[4]{162} \gt \sqrt[4]{96}$

$\therefore 3\sqrt[4]{2} \gt 2\sqrt[4]{6}$.

(ii) Given,

$\Rightarrow 7\sqrt{3} = \sqrt{7^2 \times 3} = \sqrt{147}\\[1em] \Rightarrow 3\sqrt{7} = \sqrt{3^2 \times 7} = \sqrt{63}.$

Since, 147 > 63

$\therefore \sqrt{147} \gt \sqrt{63}$

$\therefore 7\sqrt{3} \gt 3\sqrt{7}$.