1B. If for the non-zero real numbers $a, b$ and $c$ the equalities $a^{2}+a=b^{2}, b^{2}+b=c^{2}$ and $c^{2}+c=a^{2}$ hold, determine the value of the expression $(a-b)(b-c)(c-a)$.
Solution. By adding the three equations, we obtain

$$
a^{2}+b^{2}+c^{2}+a+b+c=a^{2}+b^{2}+c^{2}, \text { i.e., } a+b+c=0
$$

Accordingly, $a+b=-c, b+c=-a, c+a=-b$. From the equation $a^{2}+a=b^{2}$, it follows that $(a-b)(a+b)=-a$, and since $a+b=-c \neq 0$, we get $a-b=-\frac{a}{a+b}$. Similarly, $b-c=-\frac{b}{b+c}$ and $c-a=-\frac{c}{c+a}$. Therefore,

$$
(a-b)(b-c)(c-a)=\frac{-a}{a+b} \cdot \frac{-b}{b+c} \cdot \frac{-c}{c+a}=\frac{-a}{-c} \cdot \frac{-b}{-a} \cdot \frac{-c}{-b}=1
$$

2AB. Prove that for any real numbers $a, b$, and $c$, the equation

$$
3(a+b+c) x^{2}+4(a b+b c+c a) x+4 a b c=0
$$

has real solutions. Under what conditions are the solutions of the equation equal to each other?