Answer
$\frac{3y}{z^{4}}$
Work Step by Step
$\sqrt[3] (\frac{27y^{3}}{z^{12}})=\sqrt[3] (27y^{3})\times\sqrt[3] (\frac{1}{z^{12}})=3y\times\frac{1}{z^{4}}=\frac{3y}{z^{4}}$
We know that $\sqrt[3] (27y^{3})=3y$, because $(3y)^{3}=(3\times3\times3)\times y^{3}=27y^{3}$.
We know that $\sqrt (\frac{1}{z^{12}})=\frac{1}{z^{4}}$, because $(\frac{1}{z^{4}})^{3}=(\frac{1}{z^{4\times3}})=\frac{1}{z^{12}}$
