1-123 Let A be $4 m$, B is $\frac{1}{4}$ less than $\frac{1}{4}$ of A, C is $\frac{1}{4}$ more than $\frac{1}{4}$ of A, then the product of A, B, and C is
(A) $4 m^{3}-\frac{1}{16} m$.
(B) $4 m^{3}-\frac{1}{4} m$.
(C) $\frac{1}{16} m^{3}-\frac{1}{64} m$.
(D) $\frac{1}{4} m^{3}-\frac{1}{64}$.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] $\because$ If A is $4 m$, then B is $m-\frac{1}{4}$, and C is $m+\frac{1}{4}$, so the product of these three numbers is
$$
4 m\left(m-\frac{1}{4}\right)\left(m+\frac{1}{4}\right)=4 m^{3}-\frac{1}{4} m .
$$

Therefore, the correct choice is (B).