Chapter 4 - Exponential and Logarithmic Functions - Section 4.3 Exponential Functions - 4.3 Assess Your Understanding - Page 305: 1

$64, 4, \dfrac{1}{9}$

We expand the exponent: $4^3=4\cdot 4\cdot 4=64$ Apply the rule $a^{p/q}=(\sqrt[q]{a})^p$ to obtain: $8^{(2/3)}=(\sqrt[3] 8)^2=(\sqrt[3]{2^3})^2=2^2=4$ Apply the rule $a^{-p} = \dfrac{1}{a^p}, p \gt 0$ to obtain: $3^{-2}=\dfrac{1}{3^2}=\dfrac{1}{9}$