Intermediate Algebra, Chapter 5, 5.1, Section 5.1, Problem 126

Simplify the expression $\displaystyle \left(\dfrac{-4a^3b^2}{12a^5b^{-4}}\right)^{-3}$ so that no negative exponents appear in the final result. Assume that the variables represent nonzero real numbers.
Simplify the expression by dividing the common factor of the numerical coefficients and by subtracting the exponents of the numerator to the denominator of the same base.
$\displaystyle \left(-\dfrac{1}{3} \cdot a^{3-5} \cdot b^{2-(-4)}\right)^{-3}$
From the exponents, subtract $5$ from $3$ to get $-2$ and subtract $-4$ from $2$ to get $6$
$\displaystyle \left(-\dfrac{1}{3} \cdot a^{-2} \cdot b^{6}\right)^{-3}$
Remove the negative exponent in the numerator by rewriting $a^{−2}$ as $\dfrac{1}{a^2}$. A negative exponent follows the rule: $a^{−n}=\dfrac{1}{a^n}$

$\displaystyle \left(-\dfrac{b^{6}}{3a^2}\right)^{-3}$
Remove the negative exponent in the by rewriting $\displaystyle \left(-\dfrac{b^{6}}{3a^2}\right)^{-3}$
as $\dfrac{1}{\displaystyle \left(-\dfrac{b^{6}}{3a^2}\right)^{3} }$. A negative exponent follows the rule: $a^{−n}=\dfrac{1}{a^n}$
$\dfrac{1}{\displaystyle \left(-\dfrac{b^{6}}{3a^2}\right)^{3} }$

Take the reciprocal of the denominator and expand the exponent $3$

$\dfrac{(-1)^3(3^3)(a^2)^3}{(b^6)^3}$

Simplify
$-\dfrac{27a^6}{b^{18}}$