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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.

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    $\begingroup$ 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 $\frac{4ab}{2}$.) $\endgroup$ – Minus One-Twelfth Feb 27 at 0:41
  • $\begingroup$ WOuld it be correct to say that it equals $\frac{1}{4ab^2}$ ? $\endgroup$ – Doug Fir Feb 27 at 0:51
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    $\begingroup$ @MinusOne-Twelfth $(4ab)^{-2}$ IS NOT = $4ab/2$ $\endgroup$ – NoChance Feb 27 at 0:56
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    $\begingroup$ @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! $\endgroup$ – Billy Feb 27 at 0:58
  • $\begingroup$ @NoChance I know – that's what I said. $\endgroup$ – Minus One-Twelfth Feb 27 at 0:59
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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

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

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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}

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