Solution 3.1:4b

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By writing <math>0\textrm{.}027</math> as <math>27\cdot 10^{-3}</math>, where
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<math>27 = 3\cdot 3\cdot 3 = 3^3</math> and <math>10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3</math> we see that
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{{Displayed math||<math>\begin{align}
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\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]
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&= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,}
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\end{align}</math>}}
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where we have used <math>\sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}</math>

Current revision

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{.}

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