Find at least one non-zero polynomial $P(x, y, z)$ such that $P(a, b, c)=0$ for every three real numbers that satisfy $\sqrt[3]{a}+\sqrt[3]{b}=$ $\sqrt[3]{c}$

Remark. Polynomial in three variables refers to any expression built from $x, y, z$ and numerlas using only addition, subtraction, and multiplication. Parentheses or positive integer exponents, as in $x(y+z)^{2}$ are allowed since this can be expanded to $x y y+2 x y z+x z z$.
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.