Given x = 222, y = 223, and z = 225, evaluate the expression: x^3 + y^3 + z^3 − 3xyz Find the exact numerical value using a suitable algebraic identity for three numbers.

Introduction / Context:
This problem tests a powerful algebraic identity for three variables: x^3 + y^3 + z^3 − 3xyz. Directly computing cubes of 222, 223, and 225 would be time-consuming and error-prone. Instead, aptitude questions expect you to recognize the identity: x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). This reduces the problem to manageable squaring and multiplication. It also highlights how expressions that look massive can collapse into small numbers when the terms are close to each other (as they are here). The second factor often becomes a small value because x, y, z are near-equal, making x^2 + y^2 + z^2 close to xy + yz + zx. The skill tested is identity recognition plus careful arithmetic. The final answer is an integer and should match exactly one option.

Given Data / Assumptions:

• x = 222 • y = 223 • z = 225 • Expression: x^3 + y^3 + z^3 − 3xyz

Concept / Approach:
Use the identity: x^3 + y^3 + z^3 − 3xyz = (x + y + z)(x^2 + y^2 + z^2 − xy − yz − zx). Compute x + y + z first. Then compute the squares and pairwise products. Subtract the product-sum from the square-sum to get the second factor, and multiply to finish.

Step-by-Step Solution:
1) Compute the sum: x + y + z = 222 + 223 + 225 = 670 2) Compute squares: 222^2 = 49284 223^2 = 49729 225^2 = 50625 x^2 + y^2 + z^2 = 49284 + 49729 + 50625 = 149638 3) Compute pairwise products: xy = 222*223 = 49506 yz = 223*225 = 50175 zx = 225*222 = 49950 xy + yz + zx = 49506 + 50175 + 49950 = 149631 4) Compute the second factor: x^2 + y^2 + z^2 − xy − yz − zx = 149638 − 149631 = 7 5) Multiply: (x + y + z)*7 = 670*7 = 4690

Verification / Alternative check:
Notice the second factor became 7, a small number, confirming we correctly used the identity and the near-equality of x, y, z. A rough sense check: the expression measures how far the triple is from being perfectly symmetric (where the term might be 0 if x = y = z). Since the numbers are close, a relatively small answer like 4690 is reasonable compared to the enormous cubes involved.

Why Other Options Are Wrong:
• 4590 or 4390: typically come from arithmetic slips in xy, yz, zx. • 4950 or 4960: usually result from an incorrect second factor (using 8 instead of 7). • Any mismatch indicates the identity arithmetic was not handled consistently.

Common Pitfalls:
• Attempting direct cubing and making large-number errors. • Forgetting one product term in (xy + yz + zx). • Mistakes in squaring (especially 223^2) and carrying sums.

Final Answer:
4690