1. Given that $a$, $b$, and $c$ are pairwise distinct real numbers. Then the roots of the equation $(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ are ( ).
(A) must have two distinct real roots
(B) have no real roots
(C) must have two equal real roots
(D) the roots of the equation may take the values $a$, $b$, $c$
1.A.
The original equation can be transformed into
$$
3 x^{2}-2(a+b+c) x+a b+b c+c a=0 .
$$

Its discriminant is
$$
\begin{array}{l}
\Delta=4(a+b+c)^{2}-4 \times 3(a b+b c+c a) \\
=2\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] .
\end{array}
$$

Since $a, b, c$ are pairwise unequal, then $\Delta>0$, so the equation must have two distinct real roots.