If a = 101, evaluate the expression: a(a^2 − 3a + 3) Choose the correct value.

Introduction / Context:
This question tests algebraic expansion and recognition of a well-known binomial pattern. The expression a(a^2 − 3a + 3) can be rewritten in terms of (a − 1)^3, making evaluation for a = 101 very fast.

Given Data / Assumptions:

• a = 101 • Expression: a(a^2 − 3a + 3) • Required: exact integer value

Concept / Approach:
First expand: a(a^2 − 3a + 3) = a^3 − 3a^2 + 3a. Now recall: (a − 1)^3 = a^3 − 3a^2 + 3a − 1. So: a^3 − 3a^2 + 3a = (a − 1)^3 + 1. Then substitute a = 101 to compute (100^3 + 1).

Step-by-Step Solution:
1) Expand the product: a(a^2 − 3a + 3) = a^3 − 3a^2 + 3a 2) Compare with (a − 1)^3: (a − 1)^3 = a^3 − 3a^2 + 3a − 1 3) Therefore: a^3 − 3a^2 + 3a = (a − 1)^3 + 1 4) Substitute a = 101: (a − 1)^3 + 1 = 100^3 + 1 5) Compute 100^3: 100^3 = 100 * 100 * 100 = 1,000,000 6) Add 1: 1,000,000 + 1 = 1,000,001

Verification / Alternative check:
Direct computation also works: 101^3 = 1,030,301; 3a^2 = 3*10,201 = 30,603; 3a = 303. Then 1,030,301 − 30,603 + 303 = 1,000,001. Same result.

Why Other Options Are Wrong:
• 1,000,000: misses the +1 from the (a − 1)^3 relationship. • 999,999: incorrect subtraction handling. • 1,010,101 and 1,030,301: common distractors from partial computations like a^3 alone.

Common Pitfalls:
• Expanding incorrectly (especially the −3a term). • Not recognizing the (a − 1)^3 + 1 pattern that makes it fast.

Final Answer:
1000001