If x + y + z = 6 and xy + yz + zx = 11, what is the value of (x - y)² + (y - z)² + (z - x)²?

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A 6
B 12
C 18
D 24

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Answer: A. 6

Answer: A) 6
Explanation: (x - y)² + (y - z)² + (z - x)² = 2(x² + y² + z²) - 2(xy + yz + zx).
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx), so 36 = x² + y² + z² + 22 and x² + y² + z² = 14.
Value = 2(14) - 2(11) = 28 - 22 = 6.

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If x + y + z = 6 and xy + yz + zx = 11, what is the value of (x - y)² + (y - z)² + (z - x)²?

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