# How can I simplify $(4ab^{-1})^{-2}$

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.

• Remember, $\color{blue}{(ABC)^n = A^n B^n C^n}$. So $(4ab^{-1})^{-2} = 4^{-2} a^{-2} \left(b^{-1}\right)^{-2}$. Can you take it from here? (Also, it would not be correct to say that $(4ab)^{-2}$ equals $\frac{4ab}{2}$.) – Minus One-Twelfth Feb 27 at 0:41
• WOuld it be correct to say that it equals $\frac{1}{4ab^2}$ ? – Doug Fir Feb 27 at 0:51
• @MinusOne-Twelfth $(4ab)^{-2}$ IS NOT = $4ab/2$ – NoChance Feb 27 at 0:56
• @DougFir Be careful with how far that power "-1" ranges. $4ab^{-1}$ means $4 \times a \times b^{-1}$, not $(4ab)^{-1}$. So $4ab^{-1} = \frac{4a}{b}$. More generally, try not to skip steps: work slowly and take it one step at a time, making sure you understand the rule you're using at each stage! – Billy Feb 27 at 0:58
• @NoChance I know – that's what I said. – Minus One-Twelfth Feb 27 at 0:59

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
$$(4ab^{-1})^{-2} = \cfrac{1}{{\left(\cfrac{4a}{b}\right)}^2} = \cfrac{1}{\cfrac{16a^2}{b^2}} = \cfrac{b^2}{16a^2} \tag{2}\label{eq2}$$
Since $$(ABC)^n=A^n B^n C^n$$ as in the comments, you have: \begin{align} (4ab^{−1})^{−2}&=4^{-2}\,a^{-2}\,\left(b^{-1}\right)^{-2}\\ &=4^{-2}\,a^{-2}\,b^{-1\cdot(-2)}\\ &=\frac{1}{4^2}\cdot\frac{1}{a^2}\cdot b^2\\ &=\frac{b^2}{16\,a^2} \end{align}