Question

Write the following in descending order:
$\sqrt[2]{2\sqrt[3]{36}},\sqrt[3]{3\sqrt[4]{412}},\sqrt[2]{2\sqrt[4]{48}}$

Answer 1

The given surds $\sqrt[2]{2\sqrt[3]{36}}$, $\sqrt[3]{3\sqrt[4]{412}}$ and $\sqrt[2]{2\sqrt[4]{48}}$can be rewritten as follows:

$\sqrt[2]{2\sqrt[3]{36}}={\left(6^{\frac{1}{3}}\right)}^{\frac{1}{2}}=6^{\frac{1}{3\times 2}}=6^{\frac{1}{6}}(\because {\left(a^x\right)}^y=a^{xy})\sqrt[3]{3\sqrt[4]{412}}={\left({12}^{\frac{1}{4}}\right)}^{\frac{1}{3}}={12}^{\frac{1}{4\times 3}}={12}^{\frac{1}{12}}\sqrt[2]{2\sqrt[4]{48}}={\left(8^{\frac{1}{4}}\right)}^{\frac{1}{2}}=8^{\frac{1}{4\times 2}}=8^{\frac{1}{8}}$

The denominators of the powers of above surds are $6,12$ and $8$ and the $LCM$ of $6,12$ and $8$ is $24$. Therefore,

$\sqrt[2]{2\sqrt[3]{36}}=6^{\frac{1}{6}}=6^{\frac{1\times 4}{6\times 4}}=6^{\frac{4}{24}}=\sqrt[24]{246^4}=\sqrt[24]{241296}\sqrt[3]{3\sqrt[4]{412}}={12}^{\frac{1}{12}}={12}^{\frac{1\times 2}{12\times 2}}={12}^{\frac{2}{24}}=\sqrt[24]{24{12}^2}=\sqrt[24]{24144}\sqrt[2]{2\sqrt[4]{48}}=8^{\frac{1}{8}}=8^{\frac{1\times 3}{8\times 3}}=8^{\frac{3}{24}}=\sqrt[24]{248^3}=\sqrt[24]{24512}$

Therefore, the surds $\sqrt[2]{2\sqrt[3]{36}}$, $\sqrt[3]{3\sqrt[4]{412}}$ and $\sqrt[2]{2\sqrt[4]{48}}$ are reduced to the surd of same order $24$.

Now, since $\sqrt[24]{241296}>\sqrt[24]{24512}>\sqrt[24]{24144}$

Hence, the descending order of the given surds is $\sqrt[2]{2\sqrt[3]{36}}>$$\sqrt[2]{2\sqrt[4]{48}}>$$\sqrt[3]{3\sqrt[4]{412}}$.