If a + b = 5 and ab = 6 for real numbers a and b, use the identity for sum of cubes to determine the exact value of a³ + b³.

Introduction / Context:
This problem uses the standard identity relating a³ + b³ to a + b and ab. It is a direct algebraic application that avoids solving for a and b individually by working with symmetric expressions instead.

Given Data / Assumptions:

• a + b = 5
• ab = 6
• a and b are real numbers
• We need to find a³ + b³

Concept / Approach:
The identity a³ + b³ = (a + b)³ − 3ab(a + b) links the sum of cubes to the sum and product of the numbers. Since both a + b and ab are given directly, we can substitute and compute a³ + b³ without ever finding a or b individually.

Step-by-Step Solution:
Use identity: a³ + b³ = (a + b)³ − 3ab(a + b) Given a + b = 5, ab = 6 Compute (a + b)³ = 5³ = 125 Compute 3ab(a + b) = 3 * 6 * 5 = 90 Now a³ + b³ = 125 − 90 = 35

Verification / Alternative check:
We can confirm by solving for a and b. They satisfy t² − 5t + 6 = 0 which factors to (t − 2)(t − 3) = 0. So a and b are 2 and 3 in some order. Then a³ + b³ = 2³ + 3³ = 8 + 27 = 35, which matches the identity based result and confirms correctness.

Why Other Options Are Wrong:
32 is close but corresponds to subtracting 93 instead of 90. 38 arises from subtracting only 87. The values 34 and 30 are also off by specific margins, showing incorrect application of the identity or wrong cube of the sum. Only 35 matches both the identity and the direct substitution check.

Common Pitfalls:
Some learners confuse the formula for a³ + b³ with that for a³ − b³, or forget the factor 3 in 3ab(a + b). Others cube a + b correctly but mistakenly use ab² or a²b in place of ab. Careful recall and substitution into the correct identity prevents these mistakes.

Final Answer:
Hence, the value of a³ + b³ is 35.