4. Given that $a, b, c$ are distinct real numbers, and
$$
\begin{array}{l}
L=\frac{1}{(a-b)(a-c)}, M=\frac{1}{(b-a)(b-c)}, \\
N=\frac{1}{(c-a)(c-b)} .
\end{array}
$$

Then among the following conclusions:
(1) $a L+b M+c N=0$,
(2) $a^{2} L+b^{2} M+c^{2} N=1$,
(3) $a^{3} L+b^{3} M+c^{3} N=a+b+c$, the number of correct conclusions is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
4.D.

Let $k=\frac{1}{(a-b)(b-c)(c-a)}$, then $L=k(c-b), M=k(a-c), N=k(b-a)$.
$$
\begin{array}{l}
\text { (1) } a L+b M+c N=k[a(c-b)+b(a-c)+c(b-a)] \\
=k \times 0=0 ;
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } a^{2} L+b^{2} M+c^{2} N=k\left[a^{2}(c-b)+b^{2}(a-c)+c^{2}(b-a)\right] \\
=k \times \frac{1}{k}=1 ;
\end{array}
$$
$$
\begin{array}{l}
\text { (3) } a^{3} L+b^{3} M+c^{3} N=k\left[a^{3}(c-b)+b^{3}(a-c)+c^{3}(b-a)\right] \\
=k \times \frac{a+b+c}{k}=a+b+c .
\end{array}
$$