## Intermediate Algebra (6th Edition)

$\frac{\sqrt[3] 64}{\sqrt 64}=\frac{4}{8}=\frac{1}{2}$
We are given that $\frac{\sqrt[3] 64}{\sqrt 64}=\sqrt[3] \frac{64}{64}=\sqrt[3] 1=1$. This simplification incorrectly uses the quotient rule. In general, we can only use the quotient rule when both radicals have the same index. In this case, we must simplify without using the quotient rule, since each radical has a different index. $\frac{\sqrt[3] 64}{\sqrt 64}=\frac{4}{8}=\frac{1}{2}$ We know that $\sqrt[3] 64=4$, because $4^{3}=64$. We also know that $\sqrt 64=8$, because $8^{2}=64$.