sum($k^3$, k=1 to n) = [n$\frac{n+1}{2}$]^2 = [sum(k)]^2. This remarkable identity means the sum of cubes equals the square of the sum of first n numbers. Proved by induction. For n=5: [5*6/2]^2 = 15^2 = 225. Verify: 1+8+27+64+125 = 225. This identity is heavily tested in JEE — if you see sum of cubes, immediately write it as [n$\frac{n+1}{2}$]^2.