For a = 25, b = 15 and c = −10, evaluate a³ + b³ + c³ − 3abc.

29250. a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 30 × (950 + 25) = 30 × 975 = 29250.

For $a = 25, b = 15$ and $c = - 10$ , evaluate $a^{3} + b^{3} + c^{3} - 3 ab c$ .

Competition practiceAlgebra

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a) = 30 \times (950 + 25) = 30 \times 975 = 29250

Answer: A) 29250

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