In 2001, the sixth question of the China Mathematical Olympiad National Training Team Selection Examination was as follows:
$$
\text { Let } F=\max _{1 \leq x \leq 3}\left|x^{3}-a x^{2}-b x-c\right| \text {. Find the minimum value of } F \text { when } a 、 b \text { 、 }
$$
$c$ take all real numbers.
Solve: First, find $F^{\prime}=\max _{-1<x \leqslant 1}\left|x^{3}-a_{1} x^{2}-b_{1} x-c_{1}\right|$ $\left(a_{1}, b_{1}, c_{1} \in \mathbf{R}\right)$'s minimum value.
$$
\begin{array}{l}
\text { Let } f(x)=x^{3}-a_{1} x^{2}-b_{1} x-c_{1} \\
(-1 \leqslant x \leqslant 1),
\end{array}
$$

It is easy to know that $4 f(1)-4 f(-1)=8-8 b_{1}$,
$$
8 f\left(\frac{1}{2}\right)-8 f\left(-\frac{1}{2}\right)=2-8 b_{1} \text {. }
$$

Then
$$
\begin{array}{l}
24 F^{\prime} \geqslant 4|f(1)|+4|f(-1)| \\
\quad+8\left|f\left(\frac{1}{2}\right)\right|+8\left|f\left(-\frac{1}{2}\right)\right| \\
\geqslant\left|4 f(1)-4 f(-1)-8 f\left(\frac{1}{2}\right)+8 f\left(-\frac{1}{2}\right)\right| \\
=6 .
\end{array}
$$

Thus, $F^{\prime} \geqslant \frac{1}{4}$.
Therefore, $\max _{1 \leqslant x \leqslant 1}\left|x^{3}-a_{1} x^{2}-b_{1} x-c_{1}\right| \geqslant \frac{1}{4}$. When $a_{1}=0, b_{1}=\frac{3}{4}, c_{1}=0$, because $-1 \leqslant x \leqslant 1$. Hence, let $x=\cos \theta$. At this time
$$
\begin{array}{l}
|f(x)|=\left|\cos ^{3} \theta-\frac{3}{4} \cos \theta\right| \\
=\frac{1}{4}|\cos 3 \theta| \leqslant \frac{1}{4} .
\end{array}
$$

Then $|f(x)|_{\text {max }}=\frac{1}{4}$, so the minimum value of $F^{\prime}$ is $\frac{1}{4}$.
Now, back to the original problem:
$$
\begin{array}{l}
F=\max _{1 \leqslant x \leqslant 3}\left|x^{3}-a x^{2}-b x-c\right| \\
=\max _{-1 \leqslant x \leqslant 1}\left|(x+2)^{3}-a(x+2)^{2}-b(x+2)-c\right| \\
=\max _{-1 \leqslant x \leqslant 1} \mid x^{3}-(a-6) x^{2}-(4 a+b-12) x \\
-(4 a+2 b+c-8) \left\lvert\, \geqslant \frac{1}{4}\right. \text {. } \\
\text { The equality holds when: }\left\{\begin{array}{l}
a-6=0, \\
4 a+b-12=\frac{3}{4}, \\
4 a+2 b+c-8=0,
\end{array}\right. \\
\end{array}
$$

That is, when $a=6, b=-\frac{45}{4}, c=\frac{13}{2}$, it is achieved.
Therefore, when $a, b, c$ take all real numbers, the minimum value of $F$ is $\frac{1}{4}$.