Solution:
Cube both sides of the condition $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{z}$:
$$\begin{gathered} x+3 \sqrt[3]{x} \sqrt[3]{x} \sqrt[3]{y}+3 \sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{y}+y=z \\ 3 \sqrt[3]{x} \sqrt[3]{x} \sqrt[3]{y}+\sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{y}=z-x-y \\ 3 \sqrt[3]{x} \sqrt[3]{y}(\sqrt[3]{x}+\sqrt[3]{y})=z-x-y \end{gathered}$$
But since $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{z}$
$$\begin{gathered} 3 \sqrt[3]{x} \sqrt[3]{y} \sqrt[3]{z}=z-x-y \\ 27 x y z=(z-x-y)^{3} \end{gathered}$$
Hence $P(x, y, z)=27 x y z-(z-x-y)^{3}$ is one such polynomial.