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Given $a + b + c = 15$ and $a^{2} + b^{2} + c^{2} = 83$ , find $a^{3} + b^{3} + c^{3} - 3 ab c$ .

Competition practiceAlgebra

A180
B1245
C360
D90

Step-by-Step Solution

ab + b c + c a = \frac{225 - 83}{2} = 71. a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a) = 15 \times 12 = 180

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Answer: A) 180

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