# How do you simplify (w^2/w^4)^-3?

We have that

${\left({w}^{2} / {w}^{4}\right)}^{-} 3 = {\left({w}^{4} / {w}^{2}\right)}^{3} = {\left({w}^{2}\right)}^{3} = {w}^{6}$

Sep 1, 2016

${w}^{6}$

#### Explanation:

There are 3 laws of indices which apply here.

${x}^{m} / {x}^{n} = {x}^{m - n} , \text{ "x^-m = 1/x^m, " } {\left({x}^{a} {y}^{b}\right)}^{m} = {x}^{a m} {y}^{b m}$

It does not matter which law you apply first.

${\left({w}^{2} / {w}^{4}\right)}^{-} 3 = {\left(\frac{1}{w} ^ 2\right)}^{-} 3 \leftarrow \text{simplify in the bracket}$

${\left(\frac{1}{w} ^ 2\right)}^{-} 3 = {\left({w}^{2}\right)}^{+ 3} \leftarrow \text{invert the fraction}$

${\left({w}^{2}\right)}^{+ 3} = {w}^{6} \text{ "larr "power law}$
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OR

${\left({w}^{2} / {w}^{4}\right)}^{-} 3 = \left({w}^{-} \frac{6}{w} ^ - 12\right) \text{ "larr "power law}$

$\left({w}^{-} \frac{6}{w} ^ - 12\right) = \left({w}^{12} / {w}^{6}\right) \text{ "larr "invert the fraction}$

${w}^{12} / {w}^{6} = {w}^{6} \text{ "larr " simplify}$