Simplify cube root of 16y^7

$\sqrt[3]{16 y^{7}}$

Step 1

Rewrite $16 y^{7}$ as ${(2 y^{2})}^{3} \cdot (2 y)$ .

Factor $8$ out of $16$ .

$\sqrt[3]{8 (2) y^{7}}$

Rewrite $8$ as $2^{3}$ .

$\sqrt[3]{2^{3} \cdot 2 y^{7}}$

Factor out $y^{6}$ .

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

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

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

Move $2$ .

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

Rewrite $2^{3} {(y^{2})}^{3}$ as ${(2 y^{2})}^{3}$ .

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

Add parentheses.

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

Step 2

Pull terms out from under the radical.

$2 y^{2} \sqrt[3]{2 y}$