Aus Online Mathematik Brückenkurs 1

By writing \displaystyle 0\textrm{.}027 as \displaystyle 27\cdot 10^{-3}, where \displaystyle 27 = 3\cdot 3\cdot 3 = 3^3 and \displaystyle 10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3 we see that

\displaystyle \begin{align} \sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt] &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} \end{align}

where we have used \displaystyle \sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}