Question

Reduce the following to surds of the same order:
$\sqrt{2},\sqrt[3]{37},\sqrt[4]{411}$ and $\sqrt[12]{121671}$

Answer 1

The given surds $\sqrt{2},\sqrt[3]{37},\sqrt[4]{411}$ and $\sqrt[12]{121671}$can be rewritten as follows:

$\sqrt{2}=2^{\frac{1}{2}}\sqrt[3]{37}=7^{\frac{1}{3}}\sqrt[4]{411}={11}^{\frac{1}{4}}\sqrt[12]{121671}={1671}^{\frac{1}{12}}$

The denominators of the powers of above surds are $2,3,4$ and $12$ and the $LCM$ of $2,3,4$ and $12$ is $12$. Therefore,

$\sqrt{2}=2^{\frac{1}{2}}=2^{\frac{1\times 6}{2\times 6}}=2^{\frac{6}{12}}=\sqrt[12]{122^6}=\sqrt[12]{1264}\sqrt[3]{37}=7^{\frac{1}{3}}=7^{\frac{1\times 4}{3\times 4}}=7^{\frac{4}{12}}=\sqrt[12]{127^4}=\sqrt[12]{122401}\sqrt[4]{411}={11}^{\frac{1}{4}}={11}^{\frac{1\times 3}{4\times 3}}={11}^{\frac{3}{12}}=\sqrt[12]{12{11}^3}=\sqrt[12]{121331}\sqrt[12]{121671}$

Hence, the surds $\sqrt{2},\sqrt[3]{37},\sqrt[4]{411}$ and $\sqrt[12]{121671}$ are reduced to the surd of same order $12$.