# How do you simplify sqrt(-2)^6?

Jan 15, 2017

${\left(\sqrt{- 2}\right)}^{6} = - 8$

#### Explanation:

Note that if $a$ is any number and $m , n$ are positive integers then:

${\left({a}^{m}\right)}^{n} = {\overbrace{\left({a}^{m}\right) \times \left({a}^{m}\right) \times \ldots \times \left({a}^{m}\right)}}^{\text{n times}}$

color(white)((a^m)^n) = overbrace(overbrace((axxaxx...xxa))^"m times"xxoverbrace((axxaxx...xxa))^"m times"xx...xxoverbrace((axxaxx...xxa))^"m times")^"n times"

$\textcolor{w h i t e}{{\left({a}^{m}\right)}^{n}} = {\overbrace{a \times a \times \ldots \times a}}^{\text{mn times}}$

$\textcolor{w h i t e}{{\left({a}^{m}\right)}^{n}} = {a}^{m n}$

So in our example we find:

${\left(\sqrt{- 2}\right)}^{6} = {\left(\sqrt{- 2}\right)}^{2 \cdot 3} = {\left({\sqrt{- 2}}^{2}\right)}^{3} = {\left(- 2\right)}^{3} = - 8$

$\textcolor{w h i t e}{}$
Footnote

I demonstrated ${\left({a}^{m}\right)}^{n} = {a}^{m n}$ above since this particular "rule" fails for negative or complex values of $a$ if $m , n$ are fractional.

For example:

$- 1 = {\left(- 1\right)}^{1} = {\left(- 1\right)}^{\frac{3}{2} \cdot \frac{2}{3}} \ne {\left({\left(- 1\right)}^{\frac{3}{2}}\right)}^{\frac{2}{3}} = {\left(- i\right)}^{\frac{2}{3}} = \frac{1}{2} - \frac{\sqrt{3}}{2} i$