Question

Compare and decide which is larger amongst $\sqrt{\sqrt[4]{48}}$ and $\sqrt[3]{3\sqrt[3]{39}}$.

Answer 1

The given surds $\sqrt[]{\sqrt[4]{48}}$ and $\sqrt[3]{3\sqrt[3]{39}}$can be rewritten as follows:

$\sqrt[]{\sqrt[4]{48}}={\left(8^{\frac{1}{4}}\right)}^{\frac{1}{2}}=8^{\frac{1}{4\times 2}}=8^{\frac{1}{8}}(\because {\left(a^x\right)}^y=a^{xy})\sqrt[3]{3\sqrt[3]{39}}={\left(9^{\frac{1}{3}}\right)}^{\frac{1}{3}}=9^{\frac{1}{3\times 3}}=9^{\frac{1}{9}}$

The denominators of the powers of above surds are $8$ and $9$ and the $LCM$ of $8$ and $9$ is $72$. Therefore,

$\sqrt[]{\sqrt[4]{48}}=8^{\frac{1}{8}}=8^{\frac{1\times 9}{8\times 9}}=8^{\frac{9}{72}}=\sqrt[72]{728^9}=\sqrt[72]{72134217728}\sqrt[3]{3\sqrt[3]{39}}=9^{\frac{1}{9}}=9^{\frac{1\times 8}{9\times 8}}=9^{\frac{8}{72}}=\sqrt[72]{729^8}=\sqrt[72]{7243046721}$

Therefore, the surds $\sqrt[]{\sqrt[4]{48}}$ and $\sqrt[3]{3\sqrt[3]{39}}$ are reduced to the surd of same order $72$.

Since $\sqrt[72]{72134217728}>\sqrt[72]{7243046721}$

Therefore, $\sqrt[]{\sqrt[4]{48}}$ is larger than $\sqrt[3]{3\sqrt[3]{39}}$.