Simplify cube root of -16x^8

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

Step 1

Rewrite $- 16 x^{8}$ as ${(- 2 x^{2})}^{3} \cdot (2 x^{2})$ .

Factor $- 8$ out of $- 16$ .

$\sqrt[3]{- 8 (2) x^{8}}$

Rewrite $- 8$ as ${(- 2)}^{3}$ .

$\sqrt[3]{{(- 2)}^{3} \cdot 2 x^{8}}$

Factor out $x^{6}$ .

$\sqrt[3]{{(- 2)}^{3} \cdot 2 (x^{6} x^{2})}$

Rewrite $x^{6}$ as ${(x^{2})}^{3}$ .

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

Move $2$ .

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

Rewrite ${(- 2)}^{3} {(x^{2})}^{3}$ as ${(- 2 x^{2})}^{3}$ .

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

Add parentheses.

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

Step 2

Pull terms out from under the radical.

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