Key idea. Replace the two variables by their sum and product, then bound the sum.

1. Let \(s=a+b\), \(p=ab\). The cube identity gives

   \[a^3+b^3=s^3-3ps=s^2.\]

   Since \(s>0\), division by s yields \(p=s(s-1)/3\).

2. The square \((a-b)^2\ge0\) says \(4p\le s^2\). Substitute the expression for p and divide by positive s:

   \[\frac{4s(s-1)}3\le s^2\quad\Longrightarrow\quad4(s-1)\le3s\quad\Longrightarrow\quad s\le4.\]

3. Positive integers a and b give integer \(s\ge2\). Only \(s=2,3,4\) remain. For \(s=2\), we get \(p=2/3\), impossible for an integer product.

4. For \(s=3\), we get \(p=2\), giving \((a,b)=(1,2)\) or \((2,1)\). For \(s=4\), we get \(p=4\), giving \((a,b)=(2,2)\).

5. Check: \(1^3+2^3=9=(1+2)^2\), and \(2^3+2^3=16=(2+2)^2\). All possibilities for the sum have been considered.

Remember. An inequality can reduce an integer equation to a short, complete list of cases.