# simplify cube root of a fraction

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?

• Not quite, You can pull out part of $m^3\cdot m$ and $p^3\cdot p$, then do a little cancelling. Additionally, you seem to think that $\sqrt[3]{x^8}=x^2$, which isn't right. – Ian Coley Dec 7 '13 at 20:46
• So it would get changed to $\sqrt[3]{\frac{25p^2}{25p^2}}$ so that $\sqrt[3]{\frac{300m^4n^8p^2}{125p^6}}$ further to $\frac{mn^2\sqrt[3]{300mn^2p^2}}{5p^2}$ – James J Dec 7 '13 at 21:01

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 $c \sqrt[3]{\frac{12}{5}c}$.