2. Find all pairs of positive integers $(a, b)$ such that $a^{3}+6 a b+1$ and $b^{3}+6 a b+1$ are both perfect cubes.

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2. Let $(a, b)$ be a pair of positive integers satisfying the conditions. Without loss of generality, assume $a \geqslant b$, then: $a^{3}<a^{3}+6 a b+1 \leqslant a^{3}+6 a^{2}+1<(a+2)^{3}$, hence $a^{3}+6 a b+1=(a+1)^{3}$, which implies $b=\frac{a+1}{2}$. Substituting this into $b^{3}+6 a b+1$, we find that $\frac{1}{8}\left(a^{3}+27 a^{2}+27 a+9\right)$ is a perfect cube, thus, $a^{3}+27 a^{2}+27 a+9$ is a perfect cube. Let $a^{3}+27 a^{2}+27 a+9=(a+t)^{3}, t \in \mathbf{N}$, expanding this shows that $3 \mid t^{3}$. Therefore, $3 \mid t$. Note that: $a^{3}<a^{3}+27 a^{2}+27 a+9<(a+9)^{3}$, hence $a^{3}+27 a^{2}+27 a+9=(a+3)^{3}$ or $(a+6)^{3}$. Solving these, we find that only $a=1$, and thus $b=1$. The required pair $(a, b)=$ (1, 1).