3. If positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a+b=c-1, \\
a^{3}+b^{3}=c^{2}-1,
\end{array}\right.
$$

then $(a, b, c)=$
3. $(2,3,6)$ or $(3,2,6)$.

From the condition $a^{3}+b^{3}=c^{2}-1$, we get
$$
(a+b)\left(a^{2}-a b+b^{2}\right)=(c+1)(c-1) \text {, }
$$

and $a+b=c-1>0$.
Thus, $a^{2}-a b+b^{2}=c+1=a+b+2$
$$
\Rightarrow a^{2}-(b+1) a+b^{2}-b-2=0 \text {. }
$$

Let $a$ be the main variable. Then
$$
\begin{array}{l}
(b+1)^{2}-4\left(b^{2}-b-2\right) \geqslant 0 \\
\Rightarrow-1 \leqslant b \leqslant 3 \Rightarrow b=1,2,3 .
\end{array}
$$

When $b=1$, $a^{2}-2 a-2=0$, there is no positive integer solution;

When $b=2$, $a^{2}-3 a=0 \Rightarrow a=3$, which satisfies the requirement;

When $b=3$, $a^{2}-4 a+4=0 \Rightarrow a=2$, which satisfies the requirement.
In summary, $(a, b, c)=(2,3,6),(3,2,6)$.