```
Given real numbers $a, b, c$ are all not equal to 0, and
\[
\begin{array}{l} 
a+b+c=m, a^{2}+b^{2}+c^{2}=\frac{m^{2}}{2} . \\
\text { Find } \frac{a(m-2 a)^{2}+b(m-2 b)^{2}+c(m-2 c)^{2}}{a b c}
\end{array}
\]
```
The three roots.
Given $a+b+c=m, a^{2}+b^{2}+c^{2}=\frac{m^{2}}{2}$, we know
$$
\begin{array}{l}
a b+b c+c a \\
=\frac{1}{2}\left((a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right)=\frac{m^{2}}{4} .
\end{array}
$$

Thus, from equation (1) we get
$$
\begin{array}{l}
t^{3}-m t^{2}+\frac{m^{2} t}{4}-a b c=0 \\
\Rightarrow a b c=\frac{1}{4} t(m-2 t)^{2} .
\end{array}
$$

Then $a b c=\frac{1}{4} a(m-2 a)^{2}$,
$$
\begin{array}{l}
a b c=\frac{1}{4} b(m-2 b)^{2}, \\
a b c=\frac{1}{4} c(m-2 c)^{2} . \\
\text { Hence } \frac{a(m-2 a)^{2}+b(m-2 b)^{2}+c(m-2 c)^{2}}{a b c} \\
=12 .
\end{array}
$$

Two, let $a, b, c$ be the roots of the equation
$$
(t-a)(t-b)(t-c)=0 \text {, }
$$

i.e., $t^{3}-(a+b+c) t^{2}+(a b+b c+c a) t-a b c$ $=0$