## Task 23/76

Determine all real number triples $\{a ; b ; c\}$ for which $a ; b ; c$ are the three solutions of the equation $x^{3}-$ $a b x^{2}+b c x-c a=0$.
If a triple $\{a ; b ; c\}$ is a solution to the given equation, then by Vieta's root theorem, we have

$$
\begin{aligned}
a+b+c & =a b \\
a b+b c+c a & =b c \\
a b c & =a c
\end{aligned}
$$

From (3) it follows (4) $a=0$ or (5) $b=1$ or (6) $c=0$. We investigate these three cases separately:

1. Case: $a=0$. Then from (1) $b=-c$, arbitrary.

2. Case: $b=1$. Then from (1) $c=-1$, from (2) $a$ arbitrary.

3. Case: $c=0$. Then from (1) $a+b=a b$, from (2) $a b=0$ so in total $a=b=0$.

Thus, the following triples satisfy the given equation:

$$
\begin{array}{rr}
\text { 1. } & \{a ; b ; c\}=\{0 ; y ;-y\} \\
\text { 2. } & \{a ; b ; c\}=\{y ; 1 ;-1\} \\
\text { 3. } & \{a ; b ; c\}=\{0 ; 0 ; 0\}
\end{array}
$$

where in each case $y$ is a real number.