Use the “sum to zero” cubes identity: If a + b + c = 8, evaluate (a − 4)^3 + (b − 3)^3 + (c − 1)^3 − 3(a − 4)(b − 3)(c − 1).

Introduction / Context:
This is a direct application of the identity p^3 + q^3 + r^3 − 3pqr = (p + q + r)(p^2 + q^2 + r^2 − pq − qr − rp). Whenever p + q + r = 0, the entire expression simplifies to 0. The shifts by 4, 3, and 1 are designed precisely to make the sum vanish.

Given Data / Assumptions:

• a + b + c = 8.
• Let p = a − 4, q = b − 3, r = c − 1.

Concept / Approach:
Compute p + q + r and use the cubes identity. If p + q + r = 0, then p^3 + q^3 + r^3 − 3pqr must be 0, regardless of the actual individual values of a, b, and c.

Step-by-Step Solution:

Verification / Alternative check:
Plug in any specific numbers satisfying a + b + c = 8, e.g., a = 4, b = 3, c = 1. Then each bracket becomes 0, making the expression 0 immediately.

Why Other Options Are Wrong:
Any nonzero value would contradict the identity when p + q + r = 0.

Common Pitfalls:
Forgetting the shift and computing cubes of a, b, c directly; or using incorrect variations of the identity.

Final Answer:
0