$1 \cdot 20$ If positive numbers $a, b, c$ satisfy the equation $a^{3}+b^{3}+c^{3}-3 a b c=0$, then
(A) $a=b=c$.
(B) $a=b \neq c$.
(C) $b=c \neq a$.
(D) $a, b, c$ are all unequal.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] $a^{3}+b^{3}+c^{3}-3 a b c$
$$
=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0
$$

Given $a, b, c>0$, then $a+b+c \neq 0$,
$$
\begin{array}{c}
\therefore \quad(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0, \\
a-b=0, \quad b-c=0, \quad c-a=0, \\
a=b=c .
\end{array}
$$

Therefore, the answer is $(A)$.