Three. (50 points) Given a positive integer $n \geqslant 3$. It is known that not all of the real numbers $x_{1}, x_{2}, \cdots, x_{n}$ are 0 and they satisfy $\sum_{k=1}^{n} x_{k}=0$. If $A=\left(\sum_{k=1}^{n} x_{k}^{3}\right)^{2}, B=\left(\sum_{k=1}^{n} x_{k}^{2}\right)^{3}$, find the maximum value of $\frac{A}{B}$.
Three, notice that $\frac{A}{B}$ is a homogeneous expression. Without loss of generality, let $\sum_{k=1} x_{k}^{2}=1$, and since $A$ is positive, we can assume $\sum_{k=1}^{n} x_{k}^{3} \geqslant 0$.

The problem is transformed into: Given a positive integer $n \geqslant 3$. Real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{k=1}^{n} x_{k}=0, \sum_{k=1}^{n} x_{k}^{2}=1$. Find the maximum value of $\sum_{k=1}^{n} x_{k}^{3}$.
Lemma: For $a, b \in \mathbf{R}_{+}, a > b, x_{k} \geqslant 0$, then
$$
\left(\sum_{k=1}^{n} x_{k}^{a}\right)^{\frac{1}{a}} \leqslant\left(\sum_{k=1}^{n} x_{k}^{b}\right)^{\frac{1}{b}}.
$$

Proof: Since the original inequality is homogeneous, without loss of generality, let
$$
\sum_{k=1}^{n} x_{k}^{b}=1.
$$

Thus, for any $k$, we have $x_{k}<1$.
Therefore, $x_{k}^{a}<x_{k}^{b}, \sum_{k=1}^{n} x_{k}^{a}<\sum_{k=1}^{n} x_{k}^{b}=1$.
The lemma is proved.
Without loss of generality, let $x_{1}, x_{2}, \cdots, x_{m}<0$,
$x_{m+1}, x_{m+2}, \cdots, x_{n} \geqslant 0$.
Clearly, $1 \leqslant m \leqslant n-1$.
Set $y_{k}=-x_{k}(1 \leqslant k \leqslant m)$.
The original problem is transformed into: Given a positive integer $n \geqslant 3$. Real numbers $y_{1}, y_{2}, \cdots, y_{m}, x_{m+1}, x_{m+2}, \cdots, x_{n}$ satisfy:
$\sum_{k=m+1}^{n} x_{k}=\sum_{k=1}^{m} y_{k}, \sum_{k=1}^{m} y_{k}^{2}+\sum_{k=m+1}^{n} x_{k}^{2}=1$.
Find the maximum value of $\sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3}$.
Guess: To maximize $\sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3}$, $x_{k}(k=m+1, \cdots, n)$ should be as concentrated as possible, and $y_{k}(k=1,2, \cdots, m)$ should be as evenly distributed as possible.
In fact, by the cubic Cauchy inequality,
$\sum_{k=1}^{m} y_{k}^{3} \sum_{k=1}^{m} y_{k}^{3} \sum_{k=1}^{m} 1 \geqslant\left(\sum_{k=1}^{m} y_{k}^{2}\right)^{3}$.
Set $\sum_{k=m+1}^{n} x_{k}^{2}=t$. Then
$\sum_{k=1}^{m} y_{k}^{2}=1-t \Rightarrow \sum_{k=1}^{m} y_{k}^{3} \geqslant \sqrt{\frac{(1-t)^{2}}{m}}$.
Notice,
$$
\begin{array}{l}
1-t=\sum_{k=1}^{m} y_{k}^{2} \geqslant \frac{1}{m}\left(\sum_{k=1}^{m} y_{k}\right)^{2} \\
=\frac{1}{m}\left(\sum_{k=1}^{m} x_{k}\right)^{2} \geqslant \frac{t}{m} \\
\Rightarrow t \leqslant \frac{m}{m+1} .
\end{array}
$$

By the lemma,
$\left(\sum_{k=m+1}^{n} x_{k}^{3}\right)^{\frac{1}{3}} \leqslant\left(\sum_{k=m+1}^{n} x_{k}^{2}\right)^{\frac{1}{2}}=\sqrt{t}$.
Thus, $\sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3} \leqslant \sqrt{t^{3}}-\sqrt{\frac{(1-t)^{2}}{m}}$.
Clearly, the right-hand side of the above expression is a monotonically increasing function of $t$, and $t \leqslant \frac{m}{m+1}$.
$$
\begin{array}{l}
\text { Then } \sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3} \leqslant \sqrt{t^{3}}-\sqrt{\frac{(1-t)^{2}}{m}} \\
\leqslant \sqrt{\left(\frac{m}{m+1}\right)^{3}}-\sqrt{\frac{1}{m}\left(\frac{1}{m+1}\right)^{3}} .
\end{array}
$$

Clearly, the right-hand side of the above expression is a monotonically increasing function of $m$, and $m \leqslant n-1$.
$$
\begin{aligned}
& \text { Hence } \sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3} \\
\leqslant & \sqrt{\left(\frac{m}{m+1}\right)^{3}}-\sqrt{\frac{1}{m}\left(\frac{1}{m+1}\right)^{3}} \\
\leqslant & \sqrt{\left(\frac{n-1}{n}\right)^{3}}-\sqrt{\frac{1}{(n-1) n^{3}}} .
\end{aligned}
$$

When $m=n-1$,
$$
\begin{array}{l}
y_{1}=y_{2}=\ldots=y_{n-1}=\sqrt{\frac{1}{n(n-1)}}, \\
x_{n}=\sqrt{\frac{n-1}{n}}
\end{array}
$$

the equality holds. At this time,
$$
\left(\sum_{k=m+1}^{n} x_{k}^{3}-\sum_{k=1}^{m} y_{k}^{3}\right)^{2}=\frac{(n-2)^{2}}{n(n-1)} .
$$

Therefore, the maximum value sought is $\frac{n-2}{\sqrt{n(n-1)}}$.