Difficulty: Easy
Correct Answer: 63
Explanation:
Introduction / Context:
This algebra problem uses the standard identity relating a³ + b³ to the sum and product of a and b. It shows how you can avoid solving for a and b individually by manipulating symmetric expressions instead. This technique is useful in many competitive exam problems involving sums and products of roots.
Given Data / Assumptions:
Concept / Approach:
We use the identity:
a³ + b³ = (a + b)³ − 3ab(a + b).
Both a + b and ab are given, so we substitute directly into this formula. This avoids the need to find a and b explicitly through solving a quadratic. This method is efficient and reduces the chance of algebraic mistakes.
Step-by-Step Solution:
Recall the identity: a³ + b³ = (a + b)³ − 3ab(a + b).
Given a + b = 3 and ab = −4.
Compute (a + b)³ = 3³ = 27.
Compute 3ab(a + b) = 3(−4)(3) = −36.
Substitute into the formula: a³ + b³ = 27 − [−36].
So a³ + b³ = 27 + 36 = 63.
Verification / Alternative check:
We can confirm by constructing the quadratic equation whose roots are a and b. Since a + b = 3 and ab = −4, they are roots of t² − 3t − 4 = 0. This factors as (t − 4)(t + 1) = 0, so a and b are 4 and −1 in some order. Then a³ + b³ = 4³ + (−1)³ = 64 − 1 = 63, which matches the identity based computation perfectly.
Why Other Options Are Wrong:
36 and 12 could appear if one incorrectly computes (a + b)² or forgets the factor 3 in 3ab(a + b). The value −15 might appear from incorrectly adding 27 and −36. The value 27 corresponds to (a + b)³ alone and ignores the product term. Only 63 is consistent with both the identity and the explicit substitution check using roots 4 and −1.
Common Pitfalls:
Common errors include using the difference of cubes formula instead of the sum of cubes, misplacing the minus sign in −3ab(a + b), or mistakenly computing (a + b)³. Some learners also try to find a and b directly without realizing that the identity gives a much quicker route. Keeping the correct identity in mind simplifies the problem greatly.
Final Answer:
Thus, the value of a³ + b³ is 63.
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