Example 1 The sum of three integers $a$, $b$, and $c$ is a multiple of 6. Then, the remainder when their sum of cubes is divided by 6 is ( ).
(A) 0
(B) 2
(C) 3
(D) uncertain
(10th "Hope Cup" National Mathematics Invitational Competition (Grade 8))
Explanation: Given that the sum of three numbers $a$, $b$, and $c$ is a multiple of 6, it is not straightforward to directly determine the remainder when $a^{3}+b^{3}+c^{3}$ is divided by 6. However, if we consider the difference between the two, we can construct:
$$
\begin{array}{l}
\left(a^{3}+b^{3}+c^{3}\right)-(a+b+c) \\
=\left(a^{3}-a\right)+\left(b^{3}-b\right)+\left(c^{3}-c\right) \\
=a(a-1)(a+1)+b(b-1)(b+1)+ \\
\quad c(c-1)(c+1),
\end{array}
$$

This transforms the problem.
Since $a$ is an integer, $a-1$, $a$, and $a+1$ are three consecutive integers.
Therefore, $(a-1) a(a+1)$ is a multiple of $2 \times 3$.
Similarly, $(b-1) b(b+1)$ and $(c-1) c(c+1)$ are also multiples of 6.

Given that $a+b+c$ is a multiple of 6, it follows that $a^{3}+b^{3}+c^{3}$ is a multiple of 6.
Thus, the correct choice is $(\mathrm{A})$.