Question
Write in descending order : $2 \sqrt[3]{6}$ and $3 \sqrt[4]{2}$

Answer

$2 \sqrt[4]{6}=\sqrt[4]{2^4 \times 6}=\sqrt[4]{96}$
$ 3 \sqrt[4]{2}=\sqrt[4]{3^4}=\sqrt[4]{162}$
Since $162>96$
$\Rightarrow \sqrt[4]{162}>\sqrt[4]{96}$
$\Rightarrow 3 \sqrt[4]{2}>2 \sqrt[4]{6}$

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