In algebra with surds, the real variables x and y satisfy the simultaneous equations x − y − √18 = −1 and x + y − 3√2 = 1. Solve this system and then find the exact value of the expression 12·x·y·(x^2 − y^2).

Introduction / Context:
This question combines solving a pair of simultaneous linear equations involving surds with evaluating a higher order algebraic expression. It is a good test of comfort with square root simplification, systems of equations, and manipulation of expressions such as x^2 − y^2. The structure is typical of aptitude and entrance examinations where multiple concepts are mixed in a single numerical problem.

Given Data / Assumptions:
- Two equations are given: x − y − √18 = −1 and x + y − 3√2 = 1.
- We must first solve for x and y as real numbers.
- Then we must evaluate 12·x·y·(x^2 − y^2).
- Standard properties of surds and algebraic identities apply.

Concept / Approach:
The method is to rewrite each equation in a cleaner form and solve for x and y using addition and subtraction. We also simplify √18 as 3√2 to keep expressions consistent. Once x and y are found, we compute x^2 − y^2 using the identity x^2 − y^2 = (x − y)(x + y). Finally, we substitute all values into the expression 12·x·y·(x^2 − y^2) to obtain the exact answer in terms of √2.

Step-by-Step Solution:
1) Simplify √18 as 3√2. The first equation becomes x − y − 3√2 = −1, so x − y = −1 + 3√2. 2) The second equation is x + y − 3√2 = 1, so x + y = 1 + 3√2. 3) Add the two new equations: (x − y) + (x + y) = (−1 + 3√2) + (1 + 3√2), so 2x = 6√2 and x = 3√2. 4) Subtract the first from the second: (x + y) − (x − y) = (1 + 3√2) − (−1 + 3√2), so 2y = 2 and y = 1. 5) Compute x^2 = (3√2)^2 = 9·2 = 18 and y^2 = 1. 6) Then x^2 − y^2 = 18 − 1 = 17. 7) Now evaluate 12·x·y·(x^2 − y^2) = 12 · (3√2) · 1 · 17 = 12 · 3 · 17 · √2 = 612√2.

Verification / Alternative check:
Substitute x = 3√2 and y = 1 back into the original equations. For the first equation, x − y − √18 becomes 3√2 − 1 − 3√2 = −1, which matches the right side. For the second equation, x + y − 3√2 becomes 3√2 + 1 − 3√2 = 1, again matching. This confirms that the solution pair is correct. Since all later calculations are straightforward substitutions and basic arithmetic, the final value 612√2 is reliable.

Why Other Options Are Wrong:
Option B (512√2) and Option E (408√2) arise from incorrect arithmetic when combining coefficients, for example by mis computing x^2 − y^2 or mis multiplying by 12. Option A (0) and Option D (1) would require either x or y or x^2 − y^2 to be zero, which is clearly not the case from the solved values. Only Option C matches the exact expression value from the correct solution of the system.

Common Pitfalls:
A common problem is failing to simplify √18 into 3√2, which makes later algebra messy and error prone. Some students also mix up addition and subtraction when combining equations, leading to incorrect values for x and y. Another frequent mistake is to forget that x^2 − y^2 can be computed quickly as (x − y)(x + y), which is simpler than squaring each term again. Careful handling of surds and consistent checking against the original equations helps avoid such slips.

Final Answer:
After solving the simultaneous equations and substituting into the expression, we obtain 12·x·y·(x^2 − y^2) = 612√2.