Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$.
1. We need to find all pairs \((a, b)\) of positive integers such that \(a^3\) is a multiple of \(b^2\) and \(b-1\) is a multiple of \(a-1\).

2. First, consider the case \(b = 1\):
   - If \(b = 1\), then \(a^3\) is trivially a multiple of \(1^2\).
   - Also, \(b-1 = 0\) and \(a-1\) divides \(0\) for any \(a\). Hence, \(a-1 \mid 0\) is always true.
   - Therefore, \((a, b) = (a, 1)\) is a solution for any positive integer \(a\).

3. Now, consider the case \(b > 1\):
   - Since \(a^3\) is a multiple of \(b^2\), we have \(b^2 \mid a^3\).
   - This implies \(b \mid a^2\). Let \(b = \frac{a^2}{d}\) for some \(d \mid a^2\).

4. Given \(b-1\) is a multiple of \(a-1\), we have:
   \[
   a-1 \mid b-1 = \frac{a^2}{d} - 1
   \]
   - This implies \(a-1 \mid a^2 - d\).

5. Note that \(a^2 \equiv 1 \pmod{a-1}\), so:
   \[
   a-1 \mid d-1
   \]
   - This means \(d \geq a\) or \(d = 1\).

6. If \(d = 1\):
   - Then \(b = \frac{a^2}{d} = a^2\).
   - We need \(a^4 = b^2 \mid a^3\), which implies \(a \mid 1\), so \(a = 1\).
   - This gives \(b = a^2 = 1\), which contradicts \(b > 1\).

7. If \(d \geq a\):
   - Then \(d = a\), which gives \(b = \frac{a^2}{d} = a\).
   - This works since \(a^3\) is a multiple of \(a^2\) and \(a-1\) divides \(a-1\).

8. Therefore, the only solutions when \(b > 1\) are \(b = a\).

Conclusion:
- The only solutions are \(b = 1\) and \(b = a\).

The final answer is \( \boxed{ (a, b) = (a, 1) } \) and \((a, a)\).