Sum of Cubes

If the task is 1^3 + 2^3 + ... + n^3, use the identity sum(i^3) = [n(n+1)/2]^2. Compute the triangular number first, then square it, taking care that intermediate multiplication can overflow even when the final presentation expects a larger type or a modulo result. For modular arithmetic, divide by two before applying the modulus only when the factor is divisible, or use a modular inverse if the modulus permits it. A loop that accumulates cubes is easy to reason about and takes O(n) time, but the formula is O(1) time and O(1) space. Test n=0, n=1, and n=2, where the result is nine. If the reported prompt instead asks for cubes of arbitrary array elements, the closed form is irrelevant; clarify that distinction before coding.