9. (16 points) Find an integer-coefficient polynomial such that $\sqrt[5]{2+\sqrt{3}}+\sqrt[5]{2-\sqrt{3}}$ is a root.

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9. Let $a=\sqrt[5]{2+\sqrt{3}}, b=\sqrt[5]{2-\sqrt{3}}, x=a+b$. Then, $a b=1, x=a+\frac{1}{a}$. We have
$$
\begin{array}{l}
\left(a+\frac{1}{a}\right)^{5}=a^{5}+\frac{1}{a^{5}}+5\left(a^{3}+\frac{1}{a^{3}}\right)+10\left(a+\frac{1}{a}\right) \\
\left(a+\frac{1}{a}\right)^{3}=a^{3}+\frac{1}{a^{3}}+3\left(a+\frac{1}{a}\right)
\end{array}
$$

Solving for $a^{3}+\frac{1}{a^{3}}$ from equation (2) and substituting it into equation (1), we get
$$
\left(a+\frac{1}{a}\right)^{5}=a^{5}+\frac{1}{a^{5}}+5\left(\left(a+\frac{1}{a}\right)^{3}-3\left(a+\frac{1}{a}\right)\right)+10\left(a+\frac{1}{a}\right) .
$$

From this and using $a^{5}+\frac{1}{a^{5}}=4$, we obtain
$$
x^{5}=4+5\left(x^{3}-3 x\right)+10 x,
$$

which simplifies to
$$
x^{5}-5 x^{3}+5 x-4=0 \text {. }
$$