Let $a, b, c, d$ be four distinct real numbers, satisfying: $a+b+c+d=3, a^{2}+$ $b^{2}+c^{2}+d^{2}=45$. Find $\frac{a^{5}}{(a-b)(a-c)(a-d)}+\frac{b^{5}}{(b-a)(b-c)(b-d)}+$ $\frac{c^{5}}{(c-a)(c-b)(c-d)}+\frac{d^{5}}{(d-a)(d-b)(d-c)}$.

Let $S_{n}=\frac{a^{n}}{(a-b)(a-c)(a-d)}+\frac{b^{n}}{(b-a)(b-c)(b-d)}+$ $\frac{c^{n}}{(c-a)(c-b)(c-d)}+\frac{d^{n}}{(d-a)(d-b)(d-c)}$. By the Lagrange interpolation formula, we know
$$
\begin{array}{c}
\frac{(x-b)(x-c)(x-d)}{(a-b)(a-c)(a-d)} a^{3}+\frac{(x-a)(x-c)(x-d)}{(b-a)(b-c)(b-d)} b^{3}+ \\
\frac{(x-a)(x-b)(x-d)}{(c-a)(c-b)(c-d)} c^{3}+\frac{(x-a)(x-b)(x-c)}{(d-a)(d-b)(d-c)} d^{3}=x^{3} .
\end{array}
$$

By comparing the coefficients of $x^{3}$ on both sides, we know $S_{3}=1$. By comparing the coefficients of $x^{2}$ on both sides, and given $a+b+c+d=3$, we get $S_{4}-3 S_{3}=0$, hence $S_{4}=3$.

From the conditions, we know $a b+a c+a d+b c+b d+c d=\frac{1}{2}\left[(a+b+c+d)^{2}-\left(a^{2}+b^{2}+\right.\right.$ $\left.\left.c^{2}+d^{2}\right)\right]=-18$. Thus, by comparing the coefficients of the $x$ term on both sides of (1), we have $\sum \frac{[-18-a(b+c+d)] a^{3}}{(a-b)(a-c)(a-d)}=$ 0. Therefore, combining $a+b+c+d=3$, we get $-18 S_{3}-S_{5}+3 S_{4}=0$, hence $S_{5}=-9$. The value of the required algebraic expression is -9.