One, (This question is worth 40 points) Given $a^{2}+b^{2}+c^{2}=1$, find the maximum value of $\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right)$.
Solution: Clearly, we only need to consider the case where $\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right)>0$. There are two scenarios:
(1) If $a^{2}-b c, b^{2}-c a, c^{2}-a b$ are all greater than 0, then
$$
\begin{aligned}
& \left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right) \leq\left(\frac{a^{2}+b^{2}+c^{2}-a b-b c-c a}{3}\right)^{3} \\
= & \left(\frac{3\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2}}{6}\right)^{3} \leq\left(\frac{1}{2}\right)^{3}=\frac{1}{8}
\end{aligned}
$$

The equality holds when $a=0, b=-\frac{1}{\sqrt{2}}, c=\frac{1}{\sqrt{2}}$.
(2) If $a^{2}-b c, b^{2}-c a, c^{2}-a b$ include two negatives and one positive, without loss of generality, assume $a^{2}-b c>0$, then
$$
\begin{aligned}
\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right) & =\left(a^{2}-b c\right)\left(c a-b^{2}\right)\left(a b-c^{2}\right) \\
& \leq\left(\frac{a^{2}-b^{2}-c^{2}+a b+a c-b c}{3}\right)^{3}
\end{aligned}
$$

And $0 \leq 2\left(a^{2}-b^{2}-c^{2}+a b+a c-b c\right)=-(a-b-c)^{2}+3\left(a^{2}+b^{2}+c^{2}\right)-4\left(b^{2}+c^{2}\right) \leq 3$ hence $\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right) \leq\left(\frac{a^{2}-b^{2}-c^{2}+a b+a c-b c}{3}\right)^{3} \leq\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$

In summary, the maximum value of the original expression is $\frac{1}{8}$, which is achieved when $a=0, b=-\frac{1}{\sqrt{2}}, c=\frac{1}{\sqrt{2}}$ and its cyclic permutations.