​ Dividing Radicals

\(\dfrac{\sqrt[3]{4r^3}}{\sqrt[3]{9r}} = \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \\ \\ \text{This is really as far as it can be simplified}\\ \text{but it's common to want the denominator to be free of radicals}\\ \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} =\dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \dfrac{\sqrt[3]{9^2r^2}}{\sqrt[3]{9^2r^2}} = \dfrac{r\sqrt[3]{324r^2}}{\sqrt[3]{9^3r^3}}=\dfrac{r\sqrt[3]{324r^2}}{9r}= \dfrac{\sqrt[3]{324r^2}}{9}\)

What about the 2? Where does that go?

eh b****y h**l I missed it.

It goes in the denominator so the denominator becomes 18 rather than 9.