Question

Write the following in ascending order:
$\sqrt{2},\sqrt[3]{33},\sqrt[6]{66}$

Answer 1

The given surds $\sqrt{2}$, $\sqrt[3]{33}$ and $\sqrt[6]{66}$can be rewritten as follows:

$\sqrt{2}=2^{\frac{1}{2}}\sqrt[3]{33}=3^{\frac{1}{3}}\sqrt[6]{66}=6^{\frac{1}{6}}$

The denominators of the powers of above surds are $2,3$ and $6$ and the $LCM$ of $2,3$ and $6$ is $6$. Therefore,

$\sqrt{2}=2^{\frac{1}{2}}=2^{\frac{1\times 3}{2\times 3}}=2^{\frac{3}{6}}=\sqrt[6]{62^3}=\sqrt[6]{68}\sqrt[3]{33}=3^{\frac{1}{3}}=3^{\frac{1\times 2}{3\times 2}}=3^{\frac{2}{6}}=\sqrt[6]{63^2}=\sqrt[6]{69}\sqrt[6]{66}$

Therefore, the surds $\sqrt{2}$, $\sqrt[3]{33}$ and $\sqrt[6]{66}$ are reduced to the surd of same order $6$.

Now, since $\sqrt[6]{66}<\sqrt[6]{68}<\sqrt[6]{69}$

Hence, the ascending order of the given surds is $\sqrt[6]{66}<$$\sqrt{2}<$$\sqrt[3]{33}$.