College Algebra (11th Edition)

$\dfrac{3}{8r}$
Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $\dfrac{3r^3-9r^2}{r^2-9}\div\dfrac{8r^3}{r+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{3r^3-9r^2}{r^2-9}\cdot\dfrac{r+3}{8r^3} \\\\= \dfrac{3r^2(r-3)}{(r+3)(r-3)}\cdot\dfrac{r+3}{8r^3} \\\\= \dfrac{3\cancel{r^2}(\cancel{r-3})}{(\cancel{r+3})(\cancel{r-3})}\cdot\dfrac{\cancel{r+3}}{8\cancel{r^2}(r)} \\\\= \dfrac{3}{8r} .\end{array}