If three real numbers a, b and c satisfy a + b + c = 9 and ab + bc + ca = 18, what is the exact value of the expression a^3 + b^3 + c^3 − 3abc?

Introduction / Context:
This problem tests knowledge of a well known symmetric identity involving cubes of three numbers. The expression a^3 + b^3 + c^3 − 3abc can be rewritten in terms of the sum a + b + c and the symmetric sum ab + bc + ca. This allows evaluation of the expression without solving for a, b, and c individually.

Given Data / Assumptions:

• a + b + c = 9.
• ab + bc + ca = 18.
• a, b, c are real numbers.
• We must compute a^3 + b^3 + c^3 − 3abc.

Concept / Approach:
Use the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca). The given sums allow us to find a^2 + b^2 + c^2 from the square of a + b + c. Once we have a^2 + b^2 + c^2, we can compute the bracketed expression and multiply by a + b + c to obtain the value of a^3 + b^3 + c^3 − 3abc.

Step-by-Step Solution:
Start with the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).We know a + b + c = 9, so (a + b + c)^2 = 9^2 = 81.Substitute ab + bc + ca = 18 into the identity: 81 = a^2 + b^2 + c^2 + 2 * 18.Compute 2 * 18 = 36, giving 81 = a^2 + b^2 + c^2 + 36.Hence a^2 + b^2 + c^2 = 81 − 36 = 45.Now use the cube identity: a^3 + b^3 + c^3 − 3abc = (a + b + c)(a^2 + b^2 + c^2 − ab − bc − ca).Compute the bracket: a^2 + b^2 + c^2 − ab − bc − ca = 45 − 18 = 27.Therefore a^3 + b^3 + c^3 − 3abc = (a + b + c) * 27 = 9 * 27 = 243.

Verification / Alternative check:
If a, b, and c were the roots of some cubic equation, their symmetric sums would appear in the coefficients, and the identity used would still hold. Plugging back our values, we can check consistency: the bracket term is 27, and multiplying by 9 clearly yields 243. No contradictions arise with the given sums, confirming the calculated value is correct.

Why Other Options Are Wrong:

• 189 and 162 would result from using incorrect values for a^2 + b^2 + c^2 or miscomputing the bracket term.
• 361 and 486 are too large and would require different sums than those given for a + b + c and ab + bc + ca.
• Only 243 fits the identity perfectly when the given numeric values are substituted.

Common Pitfalls:

• Forgetting the identity a^3 + b^3 + c^3 − 3abc and instead trying to guess values for a, b, and c.
• Mishandling the calculation of a^2 + b^2 + c^2 from (a + b + c)^2.
• Incorrect subtraction when computing 45 − 18.

Final Answer:
243