Find all pairs $(a, b)$ of positive integers such that

$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
1. Let \(a-b=t\). Due to \(a^{3}-b^{3} \geq 11 a b\) we conclude that \(a > b\) so \(t\) is a positive integer and the condition can be written as

\[
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
\]

Since

\[
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t b(b+t)+t^{3}
\]

the condition can be rewritten as

\[
(11-3 t) b(b+t) \leq t^{3} \leq (12-3 t) b(b+t)
\]

We can not have \(t \geq 4\) since in that case \(t^{3} \leq (12-3 t) b(b+t)\) is not satisfied as the right hand side is not positive. Therefore it remains to check the cases when \(t \in \{1,2,3\}\). If \(t=3\), the above condition becomes

\[
2 b(b+3) \leq 27 \leq 3 b(b+3)
\]

If \(b \geq 3\), the left hand side is greater than 27 and if \(b=1\) the right hand side is smaller than 27 so there are no solutions in these cases. If \(b=2\), we get a solution \((a, b)=(5,2)\).

If \(t \leq 2\), we have

\[
(11-3 t) b(b+t) \geq (11-6) \cdot 1 \cdot (1+1) = 10 > t^{3}
\]

so there are no solutions in this case.

In summary, the only solution is \((a, b)=(5,2)\).

Solution 2. First, from \(a^{3}-b^{3} \geq 11 a b > 0\) it follows that \(a > b\), implying that \(a-b \geq 1\). Note that

\[
a^{3}-b^{3} = (a-b)\left(a^{2}+a b+b^{2}\right) = (a-b)\left[(a-b)^{2}+3 a b\right] \geq (a-b)(1+3 a b) > 3 a b(a-b)
\]

Therefore \(12 a b \geq a^{3}-b^{3} > 3 a b(a-b)\), which implies that \(a-b < 4\) so \(a-b \in \{1,2,3\}\). We discuss three possible cases:

- \(a-b=1\)

After replacing \(a=b+1\), the condition \(a^{3}-b^{3} \geq 11 a b\) reduces to \(1 \geq 8 b^{2}+8 b\), which is not satisfied for any positive integer \(b\).

- \(a-b=2\)

After replacing \(a=b+2\), the condition \(a^{3}-b^{3} \geq 11 a b\) reduces to \(8 \geq 5 b^{2}+10 b\), which is also not satisfied for any positive integer \(b\).

- \(a-b=3\)

After replacing \(a=b+3\), the condition \(a^{3}-b^{3} \geq 11 a b\) reduces to \(27 \geq 2 b^{2}+6 b\). The last inequality holds true only for \(b=1\) and \(b=2\). For \(b=1\) we get \(a=4\) and for \(b=2\) we get \(a=5\). Direct verification shows that \(a^{3}-b^{3} \leq 12 a b\) is satisfied only for \((a, b)=(5,2)\).

In summary, \((a, b)=(5,2)\) is the only pair of positive integers satisfying all conditions of the problem.