Add or subtract terms whenever possible: cube root of (16xy^3) minus y times cube root of (128x).

Step-by-step solution

1. 1

   Understand what makes radical terms combinable. Two radical terms can be added or subtracted only when they have the same index and the same radicand after simplification. Neither term looks like the other yet, so the job is to simplify both and see whether a common radical appears.

2. 2

   Simplify the first radical. Split off the largest perfect cube: $16=8\cdot 2$ and $y^{3}$ is already a cube, so $\sqrt[3]{16xy^{3}}=\sqrt[3]{8}\cdot\sqrt[3]{y^{3}}\cdot\sqrt[3]{2x}=2y\sqrt[3]{2x}.$

3. 3

   Simplify the second radical. Here $128=64\cdot 2$ and $64=4^{3}$, so $y\sqrt[3]{128x}=y\cdot\sqrt[3]{64}\cdot\sqrt[3]{2x}=4y\sqrt[3]{2x}.$

4. 4

   Combine the like terms. Both terms now carry the same radical part $\sqrt[3]{2x}$ and the same factor $y$, so subtract the coefficients: $2y\sqrt[3]{2x}-4y\sqrt[3]{2x}=(2-4)y\sqrt[3]{2x}=-2y\sqrt[3]{2x}.$

5. 5

   Check numerically. Taking $x=1$, $y=1$: the original is $\sqrt[3]{16}-\sqrt[3]{128}\approx 2.5198-5.0397=-2.5198$, and the answer is $-2\sqrt[3]{2}\approx -2.5198$, which agrees.