Answer
$-\frac{8x^3z}{9y}$.
Work Step by Step
The given expression is
$=\frac{(-2)^3x^{4}(yz)^2}{3^2xy^3z}$
Use the law of exponents $(ab)^n=a^{n}\cdot b^n$
$=\frac{-2^3x^{4}y^2\cdot z^2}{3^2xy^3z}$
Use $2^3=8$ and $ 3^2=9$.
$=-\frac{8}{9}\cdot \frac{x^{4}}{x^1}\cdot \frac{y^{2}}{y^3}\cdot \frac{z^2}{z^1}$
Use the law of exponents $\frac{a^{m}}{a^n}=a^{m-n}$
$=-\frac{8}{9}\cdot (x^{4-1})\cdot (y^{2-3})\cdot (z^{2-1})$
Simplify.
$=-\frac{8}{9}\cdot (x^{3})\cdot (y^{-1})\cdot (z^{1})$
Use the law of exponents $a^{-n}=\frac{1}{a^{n}}$
$=-\frac{8}{9}\cdot x^{3}\cdot \frac{1}{y}\cdot z$
Simplify.
$=-\frac{8x^3z}{9y}$.
