## Precalculus (6th Edition)

$\color{blue}{\dfrac{y^2}{2}}$
Use the rule $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ to obtain: $=\dfrac{y^3}{8} \cdot \dfrac{4}{y}$ Cancel the common factors first before performing the actual multiplication to obtain: $\require{cancel} =\dfrac{\cancel{y^3}y^2}{\cancel{8}2} \cdot \dfrac{\cancel{4}}{\cancel{y}} \\=\dfrac{y^2}{2} \cdot \dfrac{1}{1} \\=\color{blue}{\dfrac{y^2}{2}}$