College Algebra 7th Edition

(a) $\frac{\sqrt{216}}{\sqrt{6}}=\sqrt{\frac{216}{6}}=\sqrt{36}=6$ (b) $\sqrt[3]{2}\sqrt[3]{32}=\sqrt[3]{64}=4$ (c) $\displaystyle \sqrt[4]{\frac{1}{4}}\sqrt[4]{\frac{1}{64}}=\sqrt[4]{\frac{1}{256}}=\frac{1}{\sqrt[4]{256}}=\frac{1}{4}$
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