Question

Reduce the following to surd of the same order:
$\sqrt[3]{32},\sqrt{2}$ and $\sqrt[5]{55}$

Answer 1

The given surds $\sqrt[3]{32},\sqrt{2},\sqrt[5]{55}$can be rewritten as follows:

$\sqrt[3]{32}=2^{\frac{1}{3}}\sqrt{2}=2^{\frac{1}{2}}\sqrt[5]{55}=5^{\frac{1}{5}}$

The denominators of the powers of above surds are $2,3$ and $5$ and the $LCM$ of $2,3$ and $5$ is $30$. Therefore,

$\sqrt[3]{32}=2^{\frac{1}{3}}=2^{\frac{1\times 10}{3\times 10}}=2^{\frac{10}{30}}=\sqrt[30]{302^{10}}=\sqrt[30]{301024}\sqrt{2}=2^{\frac{1}{2}}=2^{\frac{1\times 15}{2\times 15}}=2^{\frac{15}{30}}=\sqrt[30]{302^{15}}=\sqrt[30]{3032768}\sqrt[5]{55}=5^{\frac{1}{5}}=5^{\frac{1\times 6}{5\times 6}}=2^{\frac{6}{30}}=\sqrt[30]{305^6}=\sqrt[30]{3015625}$

Hence, the surds $\sqrt[3]{32},\sqrt{2},\sqrt[5]{55}$ are reduced to the surd of same order $30$.