18. Solve the system of equations: $\left\{\begin{array}{l}x+y+z-u=12, \\ x^{2}+y^{2}+z^{2}-u^{2}=6, \\ x^{3}+y^{3}+z^{3}+u^{3}=218, \\ x y+z u=45 .\end{array}\right.$
From the second and fourth equations, we can obtain
$$
(x+y)^{2}+(z-u)^{2}=96 \text {. }
$$

But $(x+y)+(z-u)=12$,
thus $(x+y)-(z-u)=8$.
It is easy to get $x+y=10, z-u=2$.
Therefore, $x^{\circ}+y^{\prime \prime}=(x+y)^{3}-3 x y(x+y)$
$$
\begin{array}{l}
=1000-30 x y, \\
z^{3}-u^{3}=(z-u)^{3}+3 z u(z-u)=8+6 z u
\end{array}
$$

Substituting these values into the third equation, we get
$$
992-30 x y-6 z u=216 \text {, }
$$

which simplifies to $5 x y+z u=129$.
From this and the fourth equation, we know $x y=21, z u=24$.
The solutions to the system of equations can be obtained from $\left\{\begin{array}{l}x+y=10, \\ x y=21\end{array}\right.$ and
$$
\left\{\begin{array}{l}
z-u=2, \\
z u=24
\end{array}\right. \text { . }
$$