9. The equation $9 x^{3}-3 x^{2}-3 x-1=0$ has a real root of the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a, b, c$ are positive integers. Find the value of $a+b+c$.
(1 mark)
方程 $9 x^{3}-3 x^{2}-3 x-1=0$ 有一個形如 $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$ （其中 $a \cdot b \cdot c$ 為正整數）的實根。求 $a+b+c$ 的值。
(1 分)
9. 119
9. Rewrite the equation as $10 x^{3}=x^{3}+3 x^{2}+3 x+1=(x+1)^{3}$. This gives $\frac{x+1}{x}=\sqrt[3]{10}$, or
$$
x=\frac{1}{\sqrt[3]{10}-1}=\frac{\sqrt[3]{100}+\sqrt[3]{10}+1}{10-1}=\frac{\sqrt[3]{100}+\sqrt[3]{10}+1}{9} .
$$

It follows that the answer is $100+10+9=119$.