How to simplify $(4ab^{-1})^{-2}$

algebra-precalculus

As part of a wider expression I have a component $(4ab^{-1})^{-2}$

I know that using the rules of exponents, if there was no radical within the brackets I could rewrite like this:

$\frac{(4ab)}{2}$

I also know that if the only component within the brackets were $b^{-1}$ then I could multiple to be $b^{-1 * -2}$ = $b^2$

But I cannot see how to combine these two pieces.

How can I simplify $(4ab^{-1})^{-2}$? Baby steps very much appreciated.

Best Answer

There are several ways to simplify this, but I suggest you work your way from the inside out as this seems to be the easiest way in general, and is what using brackets normally implies doing. Note that $x^{-n} = \cfrac{1}{x^n}$. As such, first we get that

$$4ab^{-1} = \cfrac{4a}{b} \tag{1}\label{eq1}$$

Next, using \eqref{eq1}, plus that $\cfrac{1}{\frac{c}{d}} = \cfrac{d}{c}$, we get that

$$(4ab^{-1})^{-2} = \cfrac{1}{{\left(\cfrac{4a}{b}\right)}^2} = \cfrac{1}{\cfrac{16a^2}{b^2}} = \cfrac{b^2}{16a^2} \tag{2}\label{eq2}$$