1. Determine all triples $(a, b, c)$ of real numbers for which

$$
\begin{aligned}
a+b+c & =1, \\
a b+b c+c a & =a b c .
\end{aligned}
$$
Solution 2. We will use the standard procedure for solving systems of equations. From the first equation, we express, for example, the "unknown" $c=1-a-b$ and substitute it into the second equation, which we will then solve with respect to the "unknown" $b$ (considering $a$ as a "parameter"). After routine transformations, we obtain the quadratic equation

$$
(a-1) b^{2}+\left(a^{2}-2 a+1\right) b+\left(a-a^{2}\right)=0 .
$$

We can see that the coefficients of this equation have a common factor of $a-1$, so we will further simplify the equation before solving it:

$$
(a-1)\left[b^{2}+(a-1) b-a\right]=0 .
$$

Since the roots of the trinomial in the square brackets are $b_{1}=1$ and $b_{2}=-a$, one of the cases must occur: $a=1, b=1$, or $b=-a$. The conclusion is the same as in the first solution.