Exercise 1 Let $a, b, c, d$ be distinct real numbers, and $a+b+c+d=3, a^{2}+b^{2}+c^{2}+d^{2}=45$. Find the value of the algebraic expression
$$
\begin{array}{l}
\frac{a^{5}}{(a-b)(a-c)(a-d)}+\frac{b^{5}}{(b-a)(b-c)(b-d)} \\
+\frac{c^{5}}{(c-a)(c-b)(c-d)}+\frac{d^{5}}{(d-a)(d-b)(d-c)} .
\end{array}
$$
Let \( S_{n}=\frac{a^{n}}{(a-b)(a-c)(a-d)} \)
\[
\begin{array}{l}
+\frac{b^{n}}{(b-a)(b-c)(b-d)} \\
+\frac{c^{n}}{(c-a)(c-b)(c-d)} \\
+\frac{d^{n}}{(d-a)(d-b)(d-c)} .
\end{array}
\]

By the Lagrange interpolation formula, we know
\[
\begin{aligned}
& \frac{(x-b)(x-c)(x-d)}{(a-b)(a-c)(a-d)} a^{3}+\frac{(x-a)(x-c)(x-d)}{(b-a)(b-c)(b-d)} b^{3} \\
+ & \frac{(x-a)(x-b)(x-d)}{(c-a)(c-b)(c-d)} c^{3}+\frac{(x-a)(x-b)(x-c)}{(d-a)(d-b)(d-c)} d^{3} \\
= & x^{3} .
\end{aligned}
\]

Comparing the coefficients of \( x^{3} \) on both sides, we find \( S_{3}=1 \); comparing the coefficients of \( x^{2} \), and using \( a+b+c+d=3 \), we get \( S_{4}-3 S_{3}=0 \), hence \( S_{4}=3 \); from the given conditions, we know
\[
\begin{aligned}
& a b+a c+a d+b c+b d+c d \\
= & \frac{1}{2}\left[(a+b+c+d)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\right]=-18 .
\end{aligned}
\]

Comparing the coefficients of \( x \) on both sides of equation (1), we have
\[
\sum \frac{[-18-a(b+c+d)] a^{3}}{(a-b)(a-c)(a-d)}=0 .
\]

Thus, combining \( a+b+c+d=3 \), we get \( -18 S_{3}-S_{5}+3 S_{4}=0 \). Therefore, \( S_{5} = -9 \).