If a = 299, b = 298, and c = 297, evaluate the expression: 2a^3 + 2b^3 + 2c^3 - 6abc What is the exact value?

Introduction / Context:
This problem tests use of a cube identity: a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca). The given expression is exactly 2 times that identity form. Using the identity avoids heavy cube calculations.

Given Data / Assumptions:

• a = 299, b = 298, c = 297
• Expression: 2a^3 + 2b^3 + 2c^3 - 6abc = 2*(a^3 + b^3 + c^3 - 3abc)

Concept / Approach:
Factor out 2, then use the identity for a^3 + b^3 + c^3 - 3abc. Because a, b, c are consecutive integers, the symmetric difference term (a^2 + b^2 + c^2 - ab - bc - ca) becomes small, making the product manageable.

Step-by-Step Solution:

Verification / Alternative check:
Because the numbers are close, the difference-based formula for S is very reliable and quick. If you compute directly with cubes, you get the same 5364 but with much heavier arithmetic.

Why Other Options Are Wrong:

Common Pitfalls:
Using the identity but forgetting to multiply by 2 at the end, or computing S as (a-b)^2 + (b-c)^2 + (c-a)^2 without halving.

Final Answer:
5364