3.23 If $a$ is a natural number, and $a^{4}-4 a^{3}+15 a^{2}-30 a+27$ is a prime number, what is this prime number?
(China Beijing Junior High School Grade 2 Mathematics Competition, 1988)
[Solution]
$$
\begin{aligned}
& a^{4}-4 a^{3}+15 a^{2}-30 a+27 \\
= & \left(a^{2}-3 a+3\right)\left(a^{2}-a+9\right) .
\end{aligned}
$$

Since $a$ is a natural number, $a^{4}-4 a^{3}+15 a^{2}-30 a+27$ is a prime number, and
$$
a^{2}-3 a+3<a^{2}-a+9 \text {. }
$$

Therefore, it can only be that

Solving gives
$$
\begin{array}{c}
a^{2}-3 a+3=1, \\
a=1, \quad a=2 .
\end{array}
$$

When $a=1$, $a^{2}-a+9=9$, which is not a prime number.
When $a=2$, $a^{2}-a+9=11$ is a prime number.
Therefore, $a^{4}-4 a^{3}+15 a^{2}-30 a+27=11$.