Three, given the quadratic function $y=a x^{2}+b x+c$ whose graph has a vertex $C$, and it intersects the $x$-axis at two distinct points $A$ and $B$.
(1) If the x-coordinate of point $C$ is 3, and the distance between points $A$ and $B$ is 8, find the roots of the equation $a x^{2}-(6 a-b) x+9 a-3 b+c=0$;
(2) If the distance from point $C$ to the $x$-axis is $k$ times the distance between points $A$ and $B$, prove that $b^{2}-4 a c=15 k^{2}$.
(1) Let point $B$ be to the left of point $A$. Then it is easy to know that the coordinates of point $A$ are (-1, 0), and the coordinates of point $B$ are (7, 0).  
We can set the quadratic function as $y=a(x+1)(x-7)$.  
Therefore, the coordinates of the vertex $C$ are $(3, -16a)$.  
$$
\therefore-\frac{b}{2 a}=3, \frac{4 a c-b^{2}}{4 a}=-16 a \text {. }
$$

Then $b=-6a, c=-7a$.  
Substituting into $a x^{2}-(6 a-b) x+9 a-3 b+c=0$ we get
$$
\begin{array}{l}
\quad a x^{2}-12 a x+20 a=0 . \\
\because \quad a \neq 0, \\
\therefore \quad x^{2}-12 x+20=0 .
\end{array}
$$

Solving, we get $x_{1}=2, x_{2}=10$.
(2) From the condition, we have $|-16 a|=8 k$,
$$
\therefore a^{2}=\frac{1}{4} k^{2} \text {. }
$$

From $\frac{4 a c-b^{2}}{4 a}=-16 a$ we get
$$
\begin{array}{l}
b^{2}-4 a c=64 a^{2} . \\
\therefore b^{2}-4 a c=16 k^{2} .
\end{array}
$$