Question

Find which is larger among $3\sqrt[3]{33}$ and $4\sqrt[4]{44}$.

Answer 1

The given surds $3\sqrt[3]{33}$ and $4\sqrt[4]{44}$ can be rewritten as follows:

$3\sqrt[3]{33}=\sqrt[3]{33\times 3^3}=\sqrt[3]{33^4}=\sqrt[3]{381}={81}^{\frac{1}{3}}4\sqrt[4]{44}=\sqrt[4]{44\times 4^4}=\sqrt[4]{44^5}=\sqrt[4]{41024}={1024}^{\frac{1}{4}}$

The denominators of the powers of above surds are $3$ and $4$ and the $LCM$ of $3$ and $4$ is $12$. Therefore,

$3\sqrt[3]{33}={81}^{\frac{1}{3}}={81}^{\frac{1\times 4}{3\times 4}}={81}^{\frac{4}{12}}=\sqrt[12]{12{81}^4}=\sqrt[12]{1243046721}4\sqrt[4]{44}={1024}^{\frac{1}{4}}={1024}^{\frac{1\times 3}{4\times 3}}={1024}^{\frac{3}{12}}=\sqrt[12]{12{\left(1024\right)}^3}=\sqrt[12]{121073741824}$

Therefore, the surds $3\sqrt[3]{33}$ and $4\sqrt[4]{44}$ are reduced to the surd of same order $12$.

Here, $\sqrt[12]{121073741824}>\sqrt[12]{1243046721}$

Hence, $4\sqrt[4]{44}$ is larger than $3\sqrt[3]{33}$.