Difficulty: Medium
Correct Answer: 666
Explanation:
Introduction / Context:
This question asks you to evaluate a^3 + b^3 + c^3 − 3abc for three specific consecutive integers. Rather than expanding each cube directly, which would be long and error prone, you can use a standard identity that expresses this combination in terms of sums and products of a, b, and c. This approach is both faster and more reliable for exam purposes.
Given Data / Assumptions:
Concept / Approach:
Use the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). Since the given numbers are consecutive, sums and products can be calculated systematically. Once you compute the sum a + b + c and the combination a^2 + b^2 + c^2 − ab − bc − ca, you simply multiply them to obtain the desired value.
Step-by-Step Solution:
Verification / Alternative check:
You can confirm the pattern by noting that a, b, and c form an arithmetic progression with common difference 1 and that a + b + c is divisible by 3. In such cases, a^3 + b^3 + c^3 − 3abc is nonzero and equal to (a + b + c) times a small integer. Here, that small integer is 3, giving 222 * 3 = 666. Any attempt to expand the cubes directly will also yield the same result, though with more work.
Why Other Options Are Wrong:
The values 365, 444, 999, and 333 do not match the identity based calculation. For example, 999 would require S3 − S2 to be 4.5 instead of 3. Only 666 arises from the exact computations of the sums and squares as shown.
Common Pitfalls:
Errors often occur when computing large squares or products manually. Careless addition or multiplication can easily lead to a wrong S2 or S3. Using the identity keeps the structure clear, but you still need to be careful with arithmetic. Writing intermediate results clearly, as done above, helps avoid mistakes.
Final Answer:
The value of a^3 + b^3 + c^3 − 3abc is 666.
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