$2 \cdot 28$ If $a-b=x \neq 0$, and $a^{3}-b^{3}=19 x^{3}$, then
(A) $a=2 x$ or $a=3 x$.
(B) $a=2 x$ or $a=-3 x$.
(C) $a=-2 x$ or $a=-3 x$.
(D) $a=-2 x$ or $a=3 x$.
(China Junior High School Mathematics Correspondence Competition, 1990)
[Solution] Let $a-b=x \neq 0$, and $a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)$, so $a^{2}+a b+b^{2}=19 x^{2}$.
Substituting $b=a-x$ into the above equation and simplifying, we get
$$
a^{2}-a x-6 x^{2}=0,
$$

Solving this, we get $a=3 x$ or $-2 x$.
Therefore, the answer is $(D)$.