For real numbers a and b, if a³ + b³ = 19 and the product ab = −6, then using identities for sums of cubes, what is the value of the sum a + b?

Introduction / Context:
This question tests the algebraic identity that relates the cube of a sum to the sum of cubes and the product term. It is a standard pattern in polynomial manipulation and factorisation used to determine sums of numbers from their cubes and product.

Given Data / Assumptions:

• a³ + b³ = 19
• ab = −6
• a and b are real numbers
• We must find S = a + b

Concept / Approach:
We use the identity (a + b)³ = a³ + b³ + 3ab(a + b). This links the sum of cubes, the product ab and the sum a + b. By substituting the given values, we form an equation in S and solve the cubic, searching for a simple real root that matches the answer options.

Step-by-Step Solution:
Let S = a + b Use identity: (a + b)³ = a³ + b³ + 3ab(a + b) So S³ = 19 + 3(−6)S S³ = 19 − 18S Rearrange to standard form: S³ + 18S − 19 = 0 Now test simple integer values from the options For S = 1: 1³ + 18(1) − 19 = 1 + 18 − 19 = 0 Therefore S = 1 is a root So a + b = 1

Verification / Alternative check:
Once we know S = 1 and ab = −6, we can set up the quadratic t² − S t + ab = 0, that is t² − t − 6 = 0. This factors to (t − 3)(t + 2) = 0 with roots 3 and −2. Then a³ + b³ = 27 − 8 = 19 which matches the given condition, confirming that S = 1 is consistent.

Why Other Options Are Wrong:
For S = 5, S³ + 18S − 19 = 125 + 90 − 19 ≠ 0. For S = 7 or S = −5 or S = −7, substituting into S³ + 18S − 19 also gives non zero results. Hence none of these candidate sums satisfy the derived cubic equation.

Common Pitfalls:
Some learners mistakenly use the identity for a³ − b³ instead of a³ + b³, leading to a wrong sign pattern. Others forget the factor 3ab(a + b) in the formula. It is also common to try to solve the cubic purely algebraically when checking integer options is faster in exam settings.

Final Answer:
Thus, the required sum is a + b = 1.