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

6.142. $x^{3}-(a+b+c) x^{2}+(ab+ac+bc) x-abc=0$.
Solution.

By verification, we find that $x_{1}=a$, since $a^{3}-a^{2}-a^{2} b-a^{2} c+ a^{2} b+a^{2} c+a b c-a b c=0$. We divide the left side of the equation by $x-a$:

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

This equation can be represented as:

$$
(x-a)\left(x^{2}-(b+c) x+b c\right)=0
$$

Solving the equation $x^{2}-(b+c) x+b c=0$, we find its two roots:

$$
x_{2,3}=\frac{b+c \pm \sqrt{(b+c)^{2}-4 b c}}{2}=\frac{b+c \pm \sqrt{b^{2}+2 b c+c^{2}-4 b c}}{2}=
$$

$$
\begin{aligned}
& =\frac{b+c \pm \sqrt{b^{2}-2 b c+c^{2}}}{2}=\frac{b+c \pm \sqrt{(b-c)^{2}}}{2}=\frac{b+c \pm(b-c)}{2} \\
& x_{2}=\frac{b+c-b+c}{2}=c, x_{3}=\frac{b+c+b-c}{2}=b
\end{aligned}
$$

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