Question

Compare and decide which is larger amongst $\sqrt[3]{3\sqrt[10]{1028}}$ and $\sqrt[7]{7\sqrt[4]{430}}$.

Answer 1

The given surds $\sqrt[3]{3\sqrt[10]{1028}}$ and $\sqrt[7]{7\sqrt[4]{430}}$can be rewritten as follows:

$\sqrt[3]{3\sqrt[10]{1028}}={\left({28}^{\frac{1}{10}}\right)}^{\frac{1}{3}}={28}^{\frac{1}{10\times 3}}={28}^{\frac{1}{30}}$ ...... $(\because {\left(a^x\right)}^y=a^{xy})$

$\sqrt[7]{7\sqrt[4]{430}}={\left({30}^{\frac{1}{4}}\right)}^{\frac{1}{7}}={30}^{\frac{1}{4\times 7}}={30}^{\frac{1}{28}}$

The denominators of the powers of above surds are $28$ and $30$ and the $LCM$ of $28$ and $30$ is $420$. Therefore,

$\sqrt[3]{3\sqrt[10]{1028}}={28}^{\frac{1}{30}}={28}^{\frac{1\times 14}{30\times 14}}={28}^{\frac{14}{420}}=\sqrt[420]{420{28}^{14}}=\sqrt[420]{420(2^2\times 7{)}^{14}}=\sqrt[420]{4202^{28}\times 7^{14}}\sqrt[7]{7\sqrt[4]{430}}={30}^{\frac{1}{28}}={30}^{\frac{1\times 15}{28\times 15}}={30}^{\frac{15}{420}}=\sqrt[420]{420{30}^{15}}=\sqrt[420]{420(5\times 6{)}^{15}}=\sqrt[420]{4205^{15}\times 6^{15}}$

Therefore, the surds $\sqrt[3]{3\sqrt[10]{1028}}$ and $\sqrt[7]{7\sqrt[4]{430}}$ are reduced to the surd of same order $420$.

Since ${30}^{15}>{28}^{14}$,

Therefore, $\sqrt[7]{7\sqrt[4]{430}}$ is larger than $\sqrt[3]{3\sqrt[10]{1028}}$.