Lösung 3.1:4b
Aus Online Mathematik Brückenkurs 1
By writing \displaystyle 0.0\text{27 } as \displaystyle \text{27}\cdot \text{1}0^{-\text{3}}, where \displaystyle \text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}} and \displaystyle 10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3} we see that
\displaystyle \begin{align}
& \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\
& =3\centerdot 0.1=0.3 \\
\end{align}
where we have used
\displaystyle \sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a