3. Let $a, b, c$ be 3 distinct real numbers, and $P(x)$ be a polynomial with real coefficients. It is known that
(1) $P(x)$ divided by $(x-a)$ gives a remainder of $a$;
(2) $P(x)$ divided by $(x-b)$ gives a remainder of $b$,
(3) $P(x)$ divided by $(x-c)$ gives a remainder of $c$.
Find the remainder when $P(x)$ is divided by $(x-a)(x-b)(x-c)$.
Given that the remainder of $P(x)$ divided by $(x-a)$ is $a$, we can assume
$$
P(x)=(x-a) Q_{1}(x)+a .
$$

Thus,
$$
P(b)=(b-a) Q_{1}(b)+a=b .
$$

So $Q_{1}(b)=1, Q_{1}(x)=(x-b) Q_{2}(x)+1$. Substituting into (1) gives
$$
P(x)=(x-a)(x-b) Q_{2}(x)+x .
$$

From $P(c)=(c-a)(c-b) Q_{2}(c)+c=c$, we get
$$
Q_{2}(c)=0, Q_{2}(x)=(x-c) Q_{3}(x) .
$$

Substituting into (2) gives
$$
\begin{aligned}
P(x) & =(x-a)(x-b)(x-c) Q_{3}(x) \\
& +x .
\end{aligned}
$$

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