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


$$
\begin{gathered}
S=\frac{b c}{(a-c)(a-b)}+\frac{a c}{(b-c)(b-a)}+\frac{a b}{(c-a)(c-b)}= \\
=\frac{b c(b-c)-a c(a-c)+a b(a-b)}{(a-b)(a-c)(b-c)}
\end{gathered}
$$

De

$$
\begin{gathered}
(a-b)(a-c)(b-c)=a^{2} b-a b^{2}+b^{2} c-b c^{2}+c^{2} a-c a^{2}= \\
b c(b-c)-a c(a-c)+a b(a-b)
\end{gathered}
$$

s így

$$
S=1
$$

(Ungár Andor, Zenta)

