5. Find all positive integers $a, b, c$ such that $2^{a!} + 2^{b!} + 2^{c!}$ is a cube of some positive integer.
5. $(a, b, c)=(1,1,2),(1,2,1),(2,1,1)$.

Obviously, when $n \geqslant 3$, we have $n!\equiv 0(\bmod 3)$.
Thus, there exists a positive integer $k$ such that $n!=3 k$.
Therefore, $2^{n!}=2^{3 k}=8^{k} \equiv 1(\bmod 7)$, and the cube of a positive integer modulo 7 can only be $0, \pm 1$.
Similarly, when $n \geqslant 3$, we have $n!\equiv 0(\bmod 6)$.
Thus, there exists a positive integer $l$ such that $n!=6 l$.
Therefore, $2^{n!}=2^{6 l}=64^{l} \equiv 1(\bmod 9)$.
Without loss of generality, assume $a \geqslant b \geqslant c$.
(1) $a \geqslant 3$ and $b \geqslant 3$.
If $c \geqslant 3$, then
$$
2^{a!}+2^{b!}+2^{c!} \equiv 3(\bmod 7),
$$

which cannot be a cube number.
Similarly,
$$
\begin{array}{l}
c=2 \text { when }, 2^{a!}+2^{b!}+2^{c!} \equiv 6(\bmod 9) ; \\
c=1 \text { when, } 2^{a!}+2^{b!}+2^{c!} \equiv 4(\bmod 7),
\end{array}
$$

neither can be a cube number.
(2) $a \geqslant 3$ and $b \leqslant 2$.
If $b=c=2$, then
$$
2^{a!}+2^{b!}+2^{c!} \equiv 2(\bmod 7),
$$

which cannot be a cube number;
If $b=2, c=1$, then
$$
2^{a!}+2^{b!}+2^{c!} \equiv 7(\bmod 9),
$$

which cannot be a cube number;
If $b=c=1$, then
$$
2^{a!}+2^{b!}+2^{c!} \equiv 5(\bmod 7),
$$

which cannot be a cube number.
(3) $2 \geqslant a \geqslant b \geqslant c$.
By calculating each case, we find that only $(a, b, c)=(2,1,1)$ satisfies the condition.
In summary, the solutions are
$$
(a, b, c)=(1,1,2),(1,2,1),(2,1,1) \text {. }
$$