Question

$$(3^{2})^{-4}; (-2y^{-1}z^{2})^{-2}$$

Answer

$$(3^2)^-4;(-(2*z^2)/y)^-2$$

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[\begin{aligned}&{({3}^{2})}^{-4}\\&{(-2\times \frac{1}{y}{z}^{2})}^{-2}\end{aligned}\]
Simplify  \(2\times \frac{1}{y}{z}^{2}\)  to  \(\frac{2{z}^{2}}{y}\).
\[\begin{aligned}&{({3}^{2})}^{-4}\\&{(-\frac{2{z}^{2}}{y})}^{-2}\end{aligned}\]