I asking this question here to just check the work and that I have simplified it correctly.
Given the formula $\Large \sqrt[3]{\frac{12m^4n^8}{5p^4}}$ We need to rationalize the denominator by multiplying by $\Large \sqrt[3]{\frac{25p^4}{25p^4}}$
Doing so would leave $\Large \frac{\sqrt[3]{300m^4n^8p^4}}{\sqrt[3]{125p^8}}$ which would further simplify to $\Large \frac{n^2\sqrt[3]{300m^4p^4}}{5p^2}$
At this point the formula would simplified as far as it can be, would this be correct?
No, if you multiply this with $ \sqrt[3]{\frac{25p^\color{red}2}{25p^\color{red}2}}$, you get $$\frac{\sqrt[3]{300m^4n^8p^2}}{\sqrt[3]{125p^6}}=\frac{\sqrt[3]{300m^4n^8p^2}}{5p^2}.$$
Given the formula $\sqrt[3]{\frac{12m^4n^8}{5p^4}}$ anohter thing, reasonable to me, you can get is $ \frac{mn^2}{p} \sqrt[3]{\frac{12mn^2}{5p}}$. Use $c=\frac{mn^2}{p}$ to compactify this to $ c \sqrt[3]{\frac{12}{5}c}$.