The sum of cubes of the first ___ consecutive natural numbers is 2025.

The sum of the cubes of the first n natural numbers follows the identity 13 + 23 + ... + n3 = [n(n+1)/2]2 — that is, the sum of cubes equals the square of the sum of the first n natural numbers.

1. Let the required count of consecutive natural numbers be n, so [n(n+1)/2]2 = 2025.

2. Take the square root of both sides (only the positive root is valid since n(n+1)/2 > 0): n(n+1)/2 = 45.

3. Multiply both sides by 2: n(n+1) = 90.

4. Rewrite as a quadratic: n2 + n − 90 = 0.

5. Solve using the quadratic formula: n = [−1 + √(1+360)]/2 = [−1+19]/2 = 9.

Cross-check: substituting n = 9 back gives 9(10)/2 = 45, and 452 = 2025, which matches the given sum exactly.