Start with a few small cases
The numbers being squared—1, 3, 6, 10—are triangular numbers, the sums of the first few positive integers.
The secret is the difference between two squares
Let Tn = 1 + 2 + ··· + n = n(n+1)/2. Compare the square of one triangular number with the previous square:
Why? Substitute the formulas:
This says something delightful: each new cube is exactly the amount needed to grow one triangular-number square into the next.
Now watch everything cancel
Add the difference-of-squares identity for k = 1, 2, ..., n:
Every intermediate square appears once positively and once negatively. They cancel, leaving only Tn² (because T0 = 0). So the sum of the cubes must equal the square of the triangular number.
A way to remember it
“The sum of the first n cubes is the square of the sum of the first n counting numbers.”
Try it yourself
What is 1³ + 2³ + ··· + 10³?
First compute 1 + 2 + ··· + 10 = 55. Then square: 55² = 3,025.
What is the difference between (1 + ··· + 9)² and (1 + ··· + 8)²?
Exactly 9³ = 729, by the difference-of-squares identity.
Related: The classical sum-of-cubes identity.