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.
$\endgroup$ 52 Answers
$\begingroup$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
$$(4ab^{-1})^{-2} = \cfrac{1}{{\left(\cfrac{4a}{b}\right)}^2} = \cfrac{1}{\cfrac{16a^2}{b^2}} = \cfrac{b^2}{16a^2} \tag{2}\label{eq2}$$
$\endgroup$ $\begingroup$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}\\[1mm] &=4^{-2}\,a^{-2}\,b^{-1\cdot(-2)}\\[1mm] &=\frac{1}{4^2}\cdot\frac{1}{a^2}\cdot b^2\\[1mm] &=\frac{b^2}{16\,a^2}. \end{align}
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