1. **State the problem:** Find the number of pairs of positive integers $(a,b)$ such that $$a^3 + b^2 = 100.$$\n\n2. **Understand the constraints:** Both $a$ and $b$ are positive integers, so $a \geq 1$ and $b \geq 1$.\n\n3. **Approach:** We will try possible values of $a$ and check if $100 - a^3$ is a perfect square (since $b^2 = 100 - a^3$).\n\n4. **Check values of $a$:**\n- For $a=1$: $a^3 = 1$, so $b^2 = 100 - 1 = 99$. 99 is not a perfect square.\n- For $a=2$: $a^3 = 8$, so $b^2 = 100 - 8 = 92$. 92 is not a perfect square.\n- For $a=3$: $a^3 = 27$, so $b^2 = 100 - 27 = 73$. 73 is not a perfect square.\n- For $a=4$: $a^3 = 64$, so $b^2 = 100 - 64 = 36$. 36 is a perfect square since $6^2=36$. So $b=6$.\n- For $a=5$: $a^3 = 125$, which is greater than 100, so no solutions for $a \geq 5$.\n\n5. **Conclusion:** The only pair is $(a,b) = (4,6)$.\n\n**Final answer:** There is exactly 1 pair of positive integers $(a,b)$ such that $a^3 + b^2 = 100$.