3. Let $x, y, z$ be positive integers, and the sets
$$
\begin{aligned}
A= & \{3(x-y)(y-z)(z-x), x y z, \\
& \left.2\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)\right\}, \\
B= & \left\{(x-y)^{3}+(y-z)^{3}+(z-x)^{3},\right. \\
& \left.x+y+z, x^{4}+y^{4}+z^{4}\right\} .
\end{aligned}
$$

If $A=B$, then $x^{3}+y^{3}+z^{3}=$ $\qquad$
3. 36.

Since $x, y, z$ are positive integers, we have $x, y, z \geqslant 1$.
Notice that,
$$
\begin{array}{l}
(x-y)^{3}+(y-z)^{3}+(z-x)^{3} \\
=3(x-y)(y-z)(z-x),
\end{array}
$$

and by the AM-GM inequality, we get
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4}>x y z, \\
2\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)>x+y+z . \\
\text { Hence } 2\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)=x^{4}+y^{4}+z^{4}, \\
x y z=x+y+z . \\
\text { Also, } 2\left(y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}\right)-\left(x^{4}+y^{4}+z^{4}\right) \\
=(x+y+z)(y+z-x)(z+x-y)(x+y-z),
\end{array}
$$

we know that $y+z-x=0$ or $z+x-y=0$ or $x+y-z=0$.
Without loss of generality, assume $x=y+z$.
Substituting into $x y z=x+y+z$ gives
$$
\begin{array}{l}
(y+z) y z=2(y+z) \Rightarrow y z=2 \\
\Rightarrow\{y, z\}=\{1,2\} \\
\Rightarrow x^{3}+y^{3}+z^{3}=36 .
\end{array}
$$
5. $\frac{13 \sqrt{2}}{24}$.

According to the triangle formation conditions, if the two longer edges are skew lines (as shown in Figure 2), $A B=2 \sqrt{2}, C D=3$, taking the midpoint $E$ of $A B$, then
$$
\begin{array}{l}
C E=D E=\sqrt{2}, \\
C E+D E=2 \sqrt{2}<C D=3 .
\end{array}
$$