In algebra, the non zero real numbers x, y, and z satisfy (1/x) + (1/y) + (1/z) = 0 and also x + y + z = 9. Using these conditions, find the exact value of x^3 + y^3 + z^3 − 3xyz.

Introduction / Context:
This question links symmetric sums of variables with their reciprocals. It is a classic application of algebraic identities for sums of cubes and relationships between coefficients and roots. Instead of requiring explicit values of x, y, and z, the problem invites you to work symbolically with identities, which is a powerful technique in higher level algebra and many competitive exams.

Given Data / Assumptions:
- x, y, and z are non zero real numbers.
- We are told that (1/x) + (1/y) + (1/z) = 0.
- We also know that x + y + z = 9.
- We must compute x^3 + y^3 + z^3 − 3xyz using these relationships.

Concept / Approach:
First, use the reciprocal condition to derive information about xy + yz + zx. Since (1/x) + (1/y) + (1/z) equals (xy + yz + zx) divided by xyz, the equality to zero forces the numerator xy + yz + zx to be zero (because xyz is non zero). Then, apply the identity x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). With x + y + z and xy + yz + zx known, we can compute x^2 + y^2 + z^2 and then the entire expression.

Step-by-Step Solution:
1) Use the reciprocal condition: (1/x) + (1/y) + (1/z) = (xy + yz + zx) / (xyz) = 0. 2) Because xyz is non zero, the only way for the fraction to be zero is if xy + yz + zx = 0. 3) We also know that x + y + z = 9. 4) Compute (x + y + z)^2: this equals x^2 + y^2 + z^2 + 2(xy + yz + zx). 5) Substitute the known values: 9^2 = x^2 + y^2 + z^2 + 2·0, so x^2 + y^2 + z^2 = 81. 6) Now apply the identity x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). 7) Substitute the values: x + y + z = 9, x^2 + y^2 + z^2 = 81, and xy + yz + zx = 0, giving 9·(81 − 0) = 9·81 = 729.

Verification / Alternative check:
You can test with a simple concrete triple that satisfies the conditions. For example, take x = 3, y = 3, and z = 3. Then x + y + z = 9, but (1/x) + (1/y) + (1/z) = 1, so this choice does not satisfy the reciprocal condition. A more suitable triple is harder to guess directly, but the algebraic derivation does not require actual values. The identity used is standard and holds for any real x, y, and z, so as long as the symmetric sums satisfy the given relations, the final result 729 is valid for all such triples.

Why Other Options Are Wrong:
Option A (6561) would correspond to squaring 81 again or mis applying the identity. Option B (361) and Option D (81) are typical distractors from mis computing x^2 + y^2 + z^2 or forgetting to multiply by x + y + z. Option E (243) appears if you mistakenly multiply 27 by 9 or mis handle cubes. Only Option C exactly matches the result of applying the correct identity with the given sums.

Common Pitfalls:
Common errors include forgetting that (1/x) + (1/y) + (1/z) simplifies to (xy + yz + zx) / (xyz), or treating xy + yz + zx as zero without checking that xyz is non zero. Some students mis recall the identity for x^3 + y^3 + z^3 − 3xyz or drop terms like xy + yz + zx prematurely. Writing out identities carefully and working step by step prevents these mistakes and strengthens algebraic fluency.

Final Answer:
Using the symmetric sums and the standard identity for cubes, the value of x^3 + y^3 + z^3 − 3xyz is 729.