Question

thu
\frac{(3z)^{2}}{z^{3}}

Explanation:

Step1: Expand the numerator

First, we expand \((3z)^2\) using the power of a product rule \((ab)^n = a^n b^n\). So \((3z)^2 = 3^2 \cdot z^2 = 9z^2\).
Now our expression becomes \(\frac{9z^2}{z^3}\).

Step2: Simplify using exponent rule

We use the quotient rule for exponents \(\frac{a^m}{a^n}=a^{m - n}\) (where \(a
eq0\) and \(m,n\) are real numbers). Here, \(a = z\), \(m = 2\), and \(n = 3\).
So \(\frac{9z^2}{z^3}=9z^{2-3}=9z^{-1}\).
We can also write this as \(\frac{9}{z}\) (since \(a^{-n}=\frac{1}{a^n}\)).

Answer:

\(\frac{9}{z}\) (or \(9z^{-1}\))