# What is the square root of 8 to the 3rd power?

Mar 26, 2015

I don't know whether you mean $\sqrt{{8}^{3}}$ or ${\sqrt{8}}^{3}$.

The good news is: it doesn't matter, they are equal.

There is only only number here, but there are several ways to write it and several ways to get to the "simplist form" of $16 \sqrt{2}$

$\sqrt{{8}^{3}} = \sqrt{8 \cdot 8 \cdot 8} = \sqrt{{8}^{2} \cdot 8} = {\sqrt{8}}^{2} \cdot \sqrt{8} = 8 \sqrt{8}$

Now, $\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$ so we can continue:

$\sqrt{{8}^{3}} = 8 \sqrt{8} = 8 \sqrt{4 \cdot 2} = 8 \sqrt{4} \cdot \sqrt{2} = 8 \cdot \left(2 \cdot \sqrt{2}\right) = \left(8 \cdot 2\right) \sqrt{2} = 16 \sqrt{2}$

OR
$\sqrt{{8}^{3}} = \sqrt{{\left({2}^{3}\right)}^{3}} = \sqrt{{2}^{3 \cdot 3}} = \sqrt{{2}^{9}} = \sqrt{{2}^{8} \cdot 2}$

$= \sqrt{{2}^{8}} \sqrt{2} = \sqrt{{\left({2}^{4}\right)}^{2}} \sqrt{2} = {2}^{4} \sqrt{2} = 16 \sqrt{2}$

OR
${\sqrt{8}}^{3} = {\sqrt{8}}^{2} \cdot \sqrt{8} = 8 \sqrt{8} = 8 \sqrt{4 \cdot 2} = 8 \sqrt{4} \cdot \sqrt{2} = 8 \cdot 2 \cdot \sqrt{2} = 16 \sqrt{2}$