Example 4 Let $a, b, c$ be 3 distinct real numbers, and the real-coefficient polynomial $p(x)$, when divided by $x-a$, $x-b$, and $x-c$, yields remainders $a$, $b$, and $c$ respectively. Find the remainder when the polynomial $p(x)$ is divided by $(x-a)(x-b)(x-c)$.
Analysis Using the method of undetermined coefficients, and imitating the solution of Example 3, it is not difficult to provide the solution to this problem. Please practice this, and here we offer another method of solution.
Solution Since the remainder of $p(x)$ divided by $x-a$ is $a$, we can set $p(x)=(x-a) Q_{1}(x)+a$
Thus, $p(b)=(b-a) Q_{1}(b)+a=b$, so $Q_{1}(b)=1$. Hence, we can set $Q_{1}(x)=(x-b) Q_{2}(x)+1$, substituting into (1) we get
$$
p(x)=(x-a)(x-b) Q_{2}(x)+x
$$

From $p(c)=(c-a)(c-b) Q_{2}(c)+c=c$, we get $Q_{2}(c)=0$. Hence, we can set $Q_{2}(x)=(x-c) Q_{3}(x)$, substituting into (2) we get
$$
p(x)=(x-a)(x-b)(x-c) Q_{3}(x)+x
$$

Therefore, the remainder of $p(x)$ divided by $(x-a)(x-b)(x-c)$ is $x$.