Answer
$-\displaystyle \frac{8x^{3}z}{9y}$
Work Step by Step
Apply rule $\quad a^{mn}=(a^{m})^{n}$ , and with $(-2)^{3}=-8 \text{ and } 3^{2}=9$, to obtain
$=\dfrac{-8}{9}\cdot\dfrac{x^{4}y^{2}z^{2}}{xy^{3}z}\\
=-\dfrac{8}{9}\cdot\dfrac{x^{4} }{x }\cdot\dfrac{ y^{2} }{ y^{3} }\cdot\dfrac{ z^{2}}{ z}$
Apply rule $\displaystyle \quad \frac{a^{m}}{a^{n}}=a^{m-n}$ to obtain:
$=-\displaystyle \frac{8}{9}x^{4-1}y^{2-3}z^{2-1}$
$=-\displaystyle \frac{8}{9}x^{3}y^{-1}z^{1}\quad$
Apply rule $\displaystyle \quad a^{-n}=\frac{1}{a^{n}}$ to obtain:
$= -\displaystyle \frac{8x^{3}z}{9y}$
