If a + b + c = 9, ab + bc + ca = 26, a^3 + b^3 = 91, b^3 + c^3 = 72 and c^3 + a^3 = 35, then what is the value of the product abc?

Introduction / Context:
This is a challenging algebra problem involving three variables a, b and c with several symmetric conditions. You are given their sum, pairwise sum of products, and pairwise sums of cubes. The goal is to find the triple product abc. Problems like this test deep familiarity with identities involving sums and cubes of variables and how to combine them efficiently.

Given Data / Assumptions:

• a + b + c = 9.
• ab + bc + ca = 26.
• a^3 + b^3 = 91.
• b^3 + c^3 = 72.
• c^3 + a^3 = 35.
• a, b, c are real numbers.
• We must determine abc.

Concept / Approach:
First we find the total sum a^3 + b^3 + c^3 by adding the three given pairwise cube sums. Then we apply the well known identity: a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). We already know a + b + c and ab + bc + ca, so we can compute a^2 + b^2 + c^2 from (a + b + c)^2 and substitute everything into the identity. This will give an equation involving a^3 + b^3 + c^3, abc, and known numerical constants. Solving that equation yields abc directly.

Step-by-Step Solution:
1. Sum the three given cube relations: (a^3 + b^3) + (b^3 + c^3) + (c^3 + a^3) = 91 + 72 + 35. 2. The left side becomes 2(a^3 + b^3 + c^3). 3. So 2(a^3 + b^3 + c^3) = 91 + 72 + 35 = 198. 4. Therefore a^3 + b^3 + c^3 = 198 / 2 = 99. 5. We know a + b + c = 9. 6. Compute a^2 + b^2 + c^2 using (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). 7. (a + b + c)^2 = 9^2 = 81. 8. So 81 = a^2 + b^2 + c^2 + 2 * 26 = a^2 + b^2 + c^2 + 52. 9. Hence a^2 + b^2 + c^2 = 81 − 52 = 29. 10. Now form the expression a^2 + b^2 + c^2 − ab − bc − ca. 11. This equals 29 − 26 = 3. 12. Use the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). 13. Substitute known values: left side is 99 − 3abc. 14. Right side is (a + b + c) * (a^2 + b^2 + c^2 − ab − bc − ca) = 9 * 3 = 27. 15. So we get 99 − 3abc = 27. 16. Rearrange for abc: 99 − 27 = 3abc, so 72 = 3abc. 17. Therefore abc = 72 / 3 = 24.

Verification / Alternative check:
We can check consistency by plugging abc = 24 into the identity again. Using a + b + c = 9 and ab + bc + ca = 26, we already computed a^2 + b^2 + c^2 − ab − bc − ca = 3. Then (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca) = 9 * 3 = 27. On the left side, a^3 + b^3 + c^3 − 3abc = 99 − 3 * 24 = 99 − 72 = 27. Both sides match, confirming that abc = 24 is consistent with all given conditions.

Why Other Options Are Wrong:
Options 48, 36 and 42 do not satisfy the identity when substituted. For example, if abc were 36, then a^3 + b^3 + c^3 − 3abc would be 99 − 108 = −9, which does not equal 27. Similar contradictions arise for 48 and 42, so these values cannot be correct.

Common Pitfalls:
A common mistake is to attempt to solve for a, b and c individually, which is unnecessary and very complicated. Another pitfall is misapplying or misremembering the identity for a^3 + b^3 + c^3 − 3abc, or mixing it with the formula for (a + b + c)^3. Accurate use of the correct identity and careful computation of a^2 + b^2 + c^2 avoids these issues and leads quickly to the correct product abc.

Final Answer:
The value of the product is abc = 24.