# How do you simplify root3(-16x^8)?

$- 2 {x}^{2} \sqrt[3]{2 {x}^{2}}$

#### Explanation:

We have:

$\sqrt[3]{- 16 {x}^{8}}$

I'm going to rewrite the terms under the root sign into terms of cubes:

$\sqrt[3]{{\left(- 1\right)}^{3} \left({2}^{3}\right) {2}^{1} {x}^{3} {x}^{3} {x}^{2}}$

The cube root and the cubes inside the root are inverse functions, so we simply end up with the base term. Everything else is a "remainder" and stays within the root sign:

$- 1 \left(2\right) \left(x\right) \left(x\right) \sqrt[3]{{2}^{1} {x}^{2}}$

and cleaning this up a bit:

$- 2 {x}^{2} \sqrt[3]{2 {x}^{2}}$

While there is the same term inside as outside the root sign, we can't really factor it more than where we are, so I'll leave it here.