If a³ + b³ = 341 and ab = 30 for real numbers a and b, then by using the identity for the sum of cubes, what is the value of a + b?

Introduction / Context:
This algebra question tests knowledge of the sum of cubes identity and how to use it along with the product of two numbers to find their sum. Such problems are a part of polynomial identities and symmetric functions in competitive maths exams. The key is to recognise that a³ + b³ can be expressed in terms of a + b and ab.

Given Data / Assumptions:

• a and b are real numbers.
• a³ + b³ = 341.
• ab = 30.
• We are required to find a + b.

Concept / Approach:
Use the identity for the sum of cubes: a³ + b³ = (a + b)³ − 3ab(a + b). Let S = a + b. We can then rewrite the given equation entirely in terms of S and the known product ab. This leads to a cubic equation in S that we can solve by checking the options or by factorisation. Once we find the correct S, we have the required sum of a and b.

Step-by-Step Solution:
Let S = a + b.Using the identity: a³ + b³ = (a + b)³ − 3ab(a + b).Substitute known values: 341 = S³ − 3(30)S.This becomes 341 = S³ − 90S.Rearrange to create a cubic equation: S³ − 90S − 341 = 0.Now test the given options for S: 1, 9, 7, 11, and 13.For S = 11: 11³ − 90 × 11 − 341 = 1331 − 990 − 341 = 1331 − 1331 = 0.Therefore S = 11 satisfies the equation, so a + b = 11.

Verification / Alternative check:
We can further confirm by attempting to construct explicit values of a and b. One approach is to note that a and b are roots of the quadratic t² − S t + ab = 0, that is t² − 11 t + 30 = 0. This quadratic factorises as (t − 5)(t − 6) = 0, giving a = 5, b = 6 or vice versa. Then a³ + b³ = 5³ + 6³ = 125 + 216 = 341 and ab = 30, which matches the original conditions. Hence a + b = 11 is confirmed.

Why Other Options Are Wrong:
Substituting S = 1, 7, 9, or 13 into S³ − 90S − 341 does not give zero. For instance, S = 9 gives 729 − 810 − 341, which is negative, and S = 13 gives 2197 − 1170 − 341, which is positive. These values do not satisfy the cubic equation, so they cannot be equal to a + b under the given conditions.

Common Pitfalls:
Some students misremember the identity and write a³ + b³ = (a + b)³ − ab, which is incorrect. Others forget to multiply by 3ab in the identity. Another typical mistake is to try and guess a and b directly without using the systematic approach with S and ab, which wastes time. Using the sum of cubes identity and forming a cubic equation in S is usually the fastest route.

Final Answer:
The value of a + b is 11.