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Difficulty: MediumAlgebraic Equations and Systems in Data Sufficiency

If aa and bb are real numbers, what is the value of a3−b3a^3 - b^3?

(1) a−b=4a - b = 4
(2) a2+b2=26a^2 + b^2 = 26

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. BOTH statements (1) and (2) TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

BOTH statements (1) and (2) TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
The correct answer choice states that both statements together are sufficient, but neither alone is sufficient. Statement (1) gives the linear difference between the variables, and Statement (2) gives the sum of their squares. Neither alone yields a single value for a3−b3a^3 - b^3. However, squaring Statement (1) allows us to isolate the product ab=5ab = 5. Using the standard factoring identity a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2), we can substitute the known components to find the unique value 124.

Step-by-Step Solution

1
Rephrase the target expression using algebraic factoring identities.
a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Expanding or factoring the expression reveals that we need the values of (a−b)(a - b), (a2+b2)(a^2 + b^2), and abab.
2
Evaluate Statement (1) independently.
a−b=4a - b = 4
Knowing a−b=4a - b = 4 leaves a2+ab+b2a^2 + ab + b^2 unknown. For example, if a=4,b=0a=4, b=0, then a3−b3=64a^3-b^3=64. If a=5,b=1a=5, b=1, then a3−b3=124a^3-b^3=124. Multiple values exist, so Statement (1) alone is insufficient.
3
Evaluate Statement (2) independently.
a2+b2=26a^2 + b^2 = 26
If a=5,b=1a=5, b=1, then a3−b3=124a^3-b^3=124. If a=26,b=0a=\sqrt{26}, b=0, then a3−b3=2626a^3-b^3=26\sqrt{26}. Multiple values exist, so Statement (2) alone is insufficient.
4
Evaluate both statements combined.
ab=5ab = 5 and a3−b3=124a^3 - b^3 = 124
From Statement (1), (a−b)2=a2−2ab+b2=42=16(a - b)^2 = a^2 - 2ab + b^2 = 4^2 = 16. Substituting Statement (2) into this equation gives 26−2ab=16  ⟹  2ab=10  ⟹  ab=526 - 2ab = 16 \implies 2ab = 10 \implies ab = 5. Now substituting all values into the factored identity gives a3−b3=4×(26+5)=4×31=124a^3 - b^3 = 4 \times (26 + 5) = 4 \times 31 = 124. This gives a single, unique value.

Key Concept

Algebraic Factoring of Difference of Cubes and System Solvability
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Topics

Algebraic Equations and Systems in Data Sufficiency

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