Three, (50 points) Find all prime numbers $p$, such that $p^{2}-$ $87 p+729$ is a perfect cube.

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When $p=2,3,5,7$, the values of $p^{2}-87 p+729$ are $559$, $477$, $319$, $169$ respectively, none of which are perfect cubes. Therefore, $p \geqslant 11$.
Let $p^{2}-87 p+729=a^{3}(a \in \mathbf{N})$.
Then $a$ is an odd number.
When $a$ is an odd number less than 17, each can be tested, and no prime $p$ that meets the condition is found.
Therefore, $a \geqslant 17$.
Notice that,
$$
p(p-87)=(a-9)\left(a^{2}+9 a+81\right) \text {. }
$$

From $a^{3}=p^{2}-87 p+729a$.
Thus, $p$ must be a factor of $a^{2}+9 a+81$.
Therefore, $a^{2}+9 a+81=k p$,
$$
p-87=k(a-9) \text {, }
$$

where $k \geqslant 2$.
Substituting equation (2) into equation (1) to eliminate $p$ yields
$$
a^{2}+9 a+81=k^{2} a+87 k-9 k^{2}
$$
$$
\begin{aligned}
\Rightarrow & k^{2}(a-9) \leqslant a^{2}+9 a-93 \\
\Rightarrow & k^{2} \leqslant \frac{a^{2}+9 a-93}{a-9}=a+18+\frac{69}{a-9} \\
& \leqslant a+18+\frac{69}{8}a$, otherwise, a contradiction is derived from equation (3), let
$$
k^{2}=a+m\left(m<27, m \in \mathbf{Z}_{+}\right) \text {. }
$$

Substituting into equation (3) yields
$$
18 a+81=m a+87 k-9 m \text {. }
$$

Clearly, 3|ma.
From equation (2), and $3 \nmid p$, it follows that $3 \nmid a$. Therefore, $31 m$. When $m=12$, from equations (4) and (5) we get
$$
(a, k)=(157,13) \text {. }
$$

At this point, $p=2011$ (a prime number).
Other cases do not meet the condition.
In summary, the $p$ that meets the condition is 2011.