Write the following in ascending order of magnitude.

$\sqrt[6]{6},\sqrt[3]{7},\sqrt[4]{8}$.

$$\begin{gathered} \sqrt[6]{6},\sqrt[3]{7},\sqrt[4]{8} \\ \sqrt[6]{6}={\left(6\right)}^{\frac{1}{6}}={\left(6\right)}^{\frac{2}{12}}={\left(6^2\right)}^{\frac{1}{12}}={\left(36\right)}^{\frac{1}{12}}...\left(1\right) \\ \sqrt[3]{7}={\left(7\right)}^{\frac{1}{3}}={\left(7\right)}^{\frac{4}{12}}={\left(7^4\right)}^{\frac{1}{12}}={\left(2401\right)}^{\frac{1}{12}}...\left(2\right) \\ \sqrt[4]{8}={\left(8\right)}^{\frac{1}{4}}={\left(8\right)}^{\frac{3}{12}}={\left(8^3\right)}^{\frac{1}{12}}={\left(512\right)}^{\frac{1}{12}}...\left(3\right) \\ \mathrm{On}\mathrm{Comparing}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ {\left(36\right)}^{\frac{1}{12}}<{\left(512\right)}^{\frac{1}{12}}<{\left(2401\right)}^{\frac{1}{12}} \\ \Rightarrow \sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7} \\ \mathrm{Hence},\sqrt[6]{6}<\sqrt[4]{8}<\sqrt[3]{7}. \end{gathered}$$