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Three numbers have $a + b + c = 9, ab + b c + c a = 23$ and $ab c = 15$ . Find $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}$ .

Competition practiceAlgebra

A259
B529
C391
D289

Step-by-Step Solution

(ab + b c + c a)^{2}

= Σ

a^{2} b^{2} + 2 ab c (a + b + c)

, so Σ

a^{2} b^{2} = 529 - 2

·15·9 = 529 − 270 = 259. (They are 1, 3, 5.)

Answer: A) 259

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