## Algebra: A Combined Approach (4th Edition)

$\sqrt[3]{128y^{10}}=4y^{3}\sqrt[3]{2y}$
$\sqrt[3]{128y^{10}}$ Take the cubic root of each factor: $\sqrt[3]{128y^{10}}=\sqrt[3]{128}\cdot\sqrt[3]{y^{10}}=...$ Rewrite $128$ as $64\cdot2$ and simplify: $...=\sqrt[3]{64\cdot2}\cdot\sqrt[3]{y^{10}}=(4\sqrt[3]{2})(y^{3}\sqrt[3]{y})=4y^{3}\sqrt[3]{2y}$