Intermediate Algebra, Chapter 5, 5.1, Section 5.1, Problem 56

Evaluate the expression $\dfrac{3^{-3}}{2^{-2}}$.
Remove the negative exponent by rewriting $3^{-3}$ as $\dfrac{1}{3^3}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{3^3}\cdot\dfrac{1}{2^{-2}}$
Cubing a number is the same as multiplying the number by itself $3$ times. In this case, $3$ cube is $27$.
$\dfrac{1}{27}\cdot\dfrac{1}{2^{-2}}$
Remove the negative exponent by rewriting $2^{-2}$ as $\dfrac{1}{2^2}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{27}\cdot\dfrac{1}{\dfrac{1}{2^{2}}}$
Remove the parentheses from the denominator.
$\dfrac{1}{27}\cdot\dfrac{1}{\dfrac{1}{4}}$
Simplify the multi-level fraction by multiplying by the reciprocal of the denominator.
$\dfrac{1}{27}\cdot 4$
Multiply $\dfrac{1}{27}$ by $4$ to get $\dfrac{4}{27}$ .
$\dfrac{4}{27}$