My text shows that
$$\left(\frac{3a^2}{4b}\right)^{-3}=\frac{1}{\left(\frac{3a^2}{4b}\right)^{3}}.$$
It also shows that
$$\frac{1}{\frac{144}{b}}=\frac{b}{144}.$$
In the first equation, it seems that $(3a^2/4b)^{-3}$ was inverted to $1/(3a^2/4b)$ with the denominator raised to the power of $3$: $1/(3a^2/4b)^3$.
In the second equation, $1/(144/b)$ is equal to $(144/b)^{-1}$ (correct?), and if I do what was done in the first equation, I get what I started with: $1/(144/b)$.
So how is it that I convert $1/(144/b)$ to $b/144$? How does that accord with what happened in the first equation?
Thank you.
-Hal
Each method you suggest leads to the equivalent result. $$\left(\frac{144}{b}\right)^{-1} = \frac 1{\frac {144}b} \cdot \frac bb = \frac b{144}$$