Answer
$-\frac{z^{2}\sqrt[3] z}{3x}$
Work Step by Step
The quotient rule holds that $\sqrt[n] (\frac{a}{b})=\frac{\sqrt[n] a}{\sqrt[n] b}$ (where $\sqrt[n] a$ and $\sqrt[n] b$ are real numbers and $\sqrt[n] b$ is nonzero).
Therefore, $-\sqrt[3] (\frac{z^{7}}{27x^{3}})=-\frac{\sqrt[3] (z^{7})}{\sqrt[3] (27x^{3})}=-\frac{\sqrt[3] (z^{6})\times\sqrt[3] z}{\sqrt[3] (27x^{3}}=-\frac{z^{2}\sqrt[3] z}{3x}$
We know that $\sqrt[3] (z^{6})=z^{2}$, because $(z^{2})^{3}=z^{2\times3}=z^{6}$. Also, we know that $\sqrt[3] (27^{3})=3x$, because $(3x)^{3}=(3\times3\times3)\times x^{1+1+1}=27x^{3}$.
