Evaluate a^3 + b^3 + c^3 − 3abc for consecutive integers Find the value when a = 225, b = 226, c = 227.

Introduction / Context:This classic identity simplifies dramatically for three consecutive integers. Recognizing the pattern avoids any need to compute large cubes.

Given Data / Assumptions:

• a = 225, b = 226, c = 227 (consecutive integers).
• Expression: a^3 + b^3 + c^3 − 3abc.

Concept / Approach:Use the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). For consecutive integers with common difference 1, this further reduces numerically to 3(a + b + c).

Step-by-Step Solution:Sum: a + b + c = 225 + 226 + 227 = 678.For three consecutive integers n − 1, n, n + 1, the expression equals 3(n − 1 + n + n + 1) = 3(3n) = 9n. With n = 226, 9n = 2034.Therefore, value = 2034.

Verification / Alternative check:Directly applying the general identity also leads to the same result; evaluating differences confirms the simplification.

Why Other Options Are Wrong:2304, 2430, and 2340 are distractors near the true value; 226 is the average, not the expression’s value.

Common Pitfalls:Attempting to cube and multiply huge numbers; forgetting the special simplification for consecutive integers.

Final Answer:2034