Solution 3.1:5b

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In order to eliminate \displaystyle \sqrt[3]{7} = 7^{1/3} from the denominator, we can multiply the top and bottom of the fraction by \displaystyle 7^{2/3}. The denominator becomes \displaystyle 7^{1/3}\cdot 7^{2/3} = 7^{1/3+2/3} = 7^1 = 7 and we get

\displaystyle \frac{1}{\sqrt[3]{7}} = \frac{1}{7^{1/3}} = \frac{1}{7^{1/3}}\cdot \frac{7^{2/3}}{7^{2/3}} = \frac{7^{2/3}}{7}\,\textrm{.}