6・69 Given $f(x)=x^{4}-4 x^{3}+(3+m) x^{2}-12 x+12$, where $m$ is a real parameter.
(1) Find all integer values of $m$ such that $f(x)-f(1-x)+4 x^{3}=0$ has at least one integer solution;
(2) Find all values of $m$ such that for every real number $x, f(x) \geqslant 0$.
［Solution］(1) From the problem statement,
$$
\begin{aligned}
0= & f(x)-f(1-x)+4 x^{3} \\
= & {\left[x^{3}-4 x^{3}+(3+m) x^{2}-12 x+12\right]-\left[(1-x)^{4}-4(1-x)^{3}\right.} \\
& \left.+(3+m)(1-x)^{2}-12(1-x)+12\right]+4 x^{3} \\
= & {\left[x^{4}-(1-x)^{4}\right]-4\left[x^{3}-(1-x)^{3}\right]+(3+m)\left[x^{2}-(1-x)^{2}\right] } \\
& -12[x-(1-x)]+4 x^{3} \\
= & \left(4 x^{3}-6 x^{2}+4 x-1\right)-4\left(2 x^{3}-3 x^{2}+3 x-1\right)+(3+m)(2 x \\
& -1)-12(2 x-1)+4 x^{3} \\
= & 6 x^{2}+2(m-13) x+12-m .
\end{aligned}
$$

Since the above quadratic equation in $x$ has at least one integer solution, its discriminant must be a perfect square, i.e.,
$$
4(m-13)^{2}-24(12-m)
$$

is a perfect square, thus we have
$$
\begin{array}{l}
(m-13)^{2}-6(12-m)=y^{2}, \text{ where } y \text{ is a non-negative integer, from which we get } \\
(m-10)^{2}-y^{2}=3, \\
(m-10+y)(m-10-y)=3 .
\end{array}
$$

Since $m-10+y \geqslant m-10-y$, we have
$$
\left\{\begin{array} { l } 
{ m - 1 0 + y = 3 , } \\
{ m - 1 0 - y = 1 ; }
\end{array} \text { or } \quad \left\{\begin{array}{l}
m-10+y=-1, \\
m-10-y=-3 .
\end{array}\right.\right.
$$

Solving these, we get
$$
m=12 \text { or } m=8 .
$$

If $m=12$, from (1) we get
$$
6 x^{2}-2 x=0,
$$

which has only one integer solution $x=0$.
If $m=8$, from (1) we get $6 x^{2}-10 x+4=0$ which has only one integer solution $x=1$.

Therefore, the values of $m$ that make $f(x)-f(1-x)+4 x^{3}=0$ have at least one integer solution are 12 or 8.
(2) $f(x)=\left(x^{2}+3\right)(x-2)^{2}+(m-4) x^{2}$
If $m<4$, then $f(2)=4(m-4)<0$, which is not valid.
If $m \geqslant 4$, it is clear that for each real number $x$, $f(x) \geqslant 0$.
Therefore, the values of $m$ are $[4, \infty)$.