For the roots $x_{1}, x_{2}, x_{3}$ of the equation $x^{3}-(a+2) x^{2}+(2 a+1) x-a=0$,

$$
\frac{2}{x_{1}}+\frac{2}{x_{2}}=\frac{3}{x_{3}}
$$

holds. Solve the equation.
Solution. It is not difficult to notice that the equation has 1 and $a$ as roots. Thus, the left side can be factored by extracting $(x-1)(x-a)$:

$$
x^{3}-(a+2) x^{2}+(2 a+1) x-a=(x-1)^{2}(x-a)
$$

Therefore, the roots of the equation are 1, 1, and $a$; since $x_{1}$ and $x_{2}$ are identical in the problem, we can distinguish two cases.

1. Case: $x_{3}=1$. In this case, according to the given relationship between the roots, $2+\frac{2}{a}=3$, which means $a=2$, and the roots of the equation are then 1, 1, and 2.
2. Case: $x_{3}=a$. The relationship between the roots in this case takes the form $2+2=\frac{3}{a}$, which means $a=\frac{3}{4}$, so the roots of the equation are: 1, 1, and $\frac{3}{4}$.