Simplify cube root of 40y^10

$\sqrt[3]{40 y^{10}}$

Step 1

Rewrite $40 y^{10}$ as ${(2 y^{3})}^{3} \cdot (5 y)$ .

Factor $8$ out of $40$ .

$\sqrt[3]{8 (5) y^{10}}$

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

$\sqrt[3]{2^{3} \cdot 5 y^{10}}$

Factor out $y^{9}$ .

$\sqrt[3]{2^{3} \cdot 5 (y^{9} y)}$

Rewrite $y^{9}$ as ${(y^{3})}^{3}$ .

$\sqrt[3]{2^{3} \cdot 5 ({(y^{3})}^{3} y)}$

Move $5$ .

$\sqrt[3]{2^{3} {(y^{3})}^{3} \cdot 5 y}$

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

$\sqrt[3]{{(2 y^{3})}^{3} \cdot 5 y}$

Add parentheses.

$\sqrt[3]{{(2 y^{3})}^{3} \cdot (5 y)}$

Step 2

Pull terms out from under the radical.

$2 y^{3} \sqrt[3]{5 y}$