Algebra 1: Common Core (15th Edition)

The error is in the first step, for $27$ should also be raised to the power 2 under the root. The correct answer is $9\sqrt[3]{y^{2}}$.
The error is in the first step, for $27$ should also be raised to the power 2 under the root. $\sqrt[n]{a^{m}}=a^{m/n}$ $(27y)^{2/3}= \sqrt[3]{(27y)^{2}}$ $= \sqrt[3]{(3^{3}y)^{2}}$ $= \sqrt[3]{(3^{6})(y^{2})}$ $= \sqrt[3]{(3^{2})^{3}(y^{2})}$ $=3^{2}\sqrt[3]{y^{2}}$ $=9\sqrt[3]{y^{2}}$