If a − b = 2 and ab = 15 for real numbers a and b, then using the identity for the difference of cubes, what is the value of a³ − b³?

Introduction / Context:
This problem uses the algebraic identity for the difference of cubes to relate a³ − b³ to the difference a − b and the product ab. By combining this identity with the square of the difference, we can find a³ − b³ without explicitly solving for a and b, which is a standard technique in algebra and aptitude exams.

Given Data / Assumptions:

• a − b = 2.
• ab = 15.
• a and b are real numbers.
• We need to find a³ − b³.

Concept / Approach:
We recall the identity: a³ − b³ = (a − b)(a² + ab + b²). We know a − b directly, but not a² + ab + b². However, we can find a² + b² by using (a − b)² = a² − 2ab + b². Once we get a² + b², we add ab to obtain a² + ab + b², and then multiply by a − b to obtain a³ − b³.

Step-by-Step Solution:
Given a − b = 2 and ab = 15. First compute (a − b)² = a² − 2ab + b². Since a − b = 2, (a − b)² = 4. So a² − 2ab + b² = 4. Substitute ab = 15: a² − 2(15) + b² = 4. Thus a² + b² − 30 = 4, so a² + b² = 34. Now compute a² + ab + b² = (a² + b²) + ab = 34 + 15 = 49. Use the difference of cubes identity: a³ − b³ = (a − b)(a² + ab + b²). So a³ − b³ = 2 · 49 = 98.

Verification / Alternative check:
We can solve for a and b explicitly to confirm. Since a − b = 2 and ab = 15, they are roots of the quadratic t² − (a + b)t + ab = 0. We can find a + b from (a − b)² = a² − 2ab + b² and (a + b)² = a² + 2ab + b². Adding these gives 2(a² + b²) = (a − b)² + (a + b)². But it is easier to note that a + b must satisfy (a + b)² = (a − b)² + 4ab = 4 + 60 = 64, so a + b = 8 or −8. With ab = 15 and a − b = 2, we find a = 5 and b = 3. Then a³ − b³ = 125 − 27 = 98, confirming the result.

Why Other Options Are Wrong:
Values like 152, 112, 108 and 90 can arise from miscomputing a² + b² or from incorrectly applying the difference of cubes formula (for example using a² − ab + b²). None of these match the exact result verified both through identities and through explicit numbers. Only 98 satisfies all the given conditions.

Common Pitfalls:
Learners sometimes confuse the sum and difference of cubes identities or forget to include the ab term when computing a² + ab + b². Mistakes in squaring a − b or in substituting ab can also lead to the wrong result. Carefully following the identity and checking with a numerical example is a good way to avoid such errors.

Final Answer:
Therefore, the value of a³ − b³ is 98.