$5 \cdot 46$ Find an integer coefficient polynomial such that $\alpha=\sqrt[3]{2}+\sqrt[3]{3}$ is its root,


[Solution]
$$
\text { Since } \begin{aligned}
\alpha^{3} & =(\sqrt[3]{2}+\sqrt[3]{3})^{3} \\
& =5+3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{3} \cdot(\sqrt[3]{2}+\sqrt[3]{3}) \\
& =5+3 \sqrt[3]{6} \alpha,
\end{aligned}
$$

thus $\left(\alpha^{3^{1}}-5\right)^{3}=(3 \sqrt[3]{6} \alpha)^{3}=162 \alpha^{3}$,
which means $\alpha^{9}-15 \alpha^{6}-87 \alpha^{3}-125=0$.
Therefore, $p(x)=x^{9}-15 x^{6}-87 x^{3}-125$ is a polynomial that satisfies the condition. Multiplying $p(x)$ by any polynomial $f(x)$, then $p(x) f(x)$ still satisfies the condition.