7. Let the quartic equation be $F(x)=f(x)-k=0$. When $k=10,20,30$, the roots of the equation $F(x)=0$ are $1,2,3$, respectively. Then the value of $f(10)+f(-6)$ is
7. Fill in 8104. Reason: According to the problem, we can consider the function $f(x)=(x-1)(x-2)(x-3)(x-m)+10 \cdot x$, thus,
$$
\begin{array}{l}
f(10)+f(-6)=[9 \cdot 8 \cdot 7 \cdot(10-m)+10 \cdot 10]+[(-7) \cdot(-8) \cdot(-9) \cdot(-6-m)-6 \cdot 10]= \\
9 \cdot 8 \cdot 7 \cdot(10-m+6-m)+100-60=8104 .
\end{array}
$$