Example 14 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=10$, $f(2)=20$, $f(3)=30$, then $f(10)+f(-6)=$ $\qquad$
(1998, Zhongshan City, Guangdong Province Mathematics Competition)
Analysis: Based on $f(1)=10, f(2)=20, f(3)=$ 30, construct
$$
\begin{array}{c}
f(x)=(x-1)(x-2)(x-3)(x-t)+10 x . \\
\text { Hence } f(10)+f(-6)=9 \times 8 \times 7(10-t)+ \\
10 \times 10+[(-7) \times(-8) \times(-9)(-6-t)- \\
60]=8104 .
\end{array}
$$