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For real a, $b, c$ with $a^{2} + b^{2} + c^{2} = ab + b c + c a$ and $a + b + c = 18$ , find $ab c$ .

Competition practiceAlgebra

A216
B18
C108
D324

Step-by-Step Solution

a^{2} + b^{2} + c^{2} - ab - b c - c a

= ½

[(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0

forces

a = b = c = 6

, so

ab c = 216

.

Answer: A) 216

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