$2.50 x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=0$.

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$2.50 x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c=0$.
2.50 It is easy to see that one of the roots of the given equation is $x_{1}=a$. Then

$$
\begin{aligned}
& {\left[x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c\right]:(x-a)=} \\
& =x^{2}-(b+c) x+b c
\end{aligned}
$$

From the equation $x^{2}-(b+c) x+b c=0$, we find that $x_{2}=b$, $x_{3}=c$.

Answer: $x_{1}=a, x_{2}=b, x_{3}=c$.

Remark. One can formulate and prove the so-called generalized Vieta's theorem. If $x_{1}, x_{2}, x_{3}$ are the roots of the equation

$$
x^{3}+p x^{2}+q x+s=0
$$

then

$$
x_{1}+x_{2}+x_{3}=-p ; x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=q ; x_{1} x_{2} x_{3}=-s
$$