Perfect cubes with one impostor — choose the odd one out Set: 1, 8, 27, 64, 124, 216, 343. Which number does not belong?

Introduction / Context:
Most terms here are perfect cubes, making the single non-cube the odd one out. Identifying cubes is straightforward by remembering small n^3 values or quickly testing cube roots for integers.

Given Data / Assumptions:

• Numbers: 1, 8, 27, 64, 124, 216, 343.
• Known cubes: 1^3=1, 2^3=8, 3^3=27, 4^3=64, 6^3=216, 7^3=343.

Concept / Approach:
Check each term for being an exact cube. The non-cube in the set is the outlier.

Step-by-Step Solution:
1 = 1^3 ⇒ cube.8 = 2^3 ⇒ cube.27 = 3^3 ⇒ cube.64 = 4^3 ⇒ cube.124 is not a perfect cube (5^3 = 125; 124 is one less).216 = 6^3 ⇒ cube.343 = 7^3 ⇒ cube.

Verification / Alternative check:
Since 124 falls directly between 5^3 − 1 and 5^3, it cannot be a cube. All other numbers are exact cubes of small integers.

Why Other Options Are Wrong:
8, 27, 64, and 216 are perfect cubes. They belong to the dominant category and thus are not the odd ones out.

Common Pitfalls:
Confusing 124 with 125 (which is 5^3). Carefully verify the exact value before concluding.

Final Answer:
124