If a + b + c = 9, and $a^{2} + b^{2} + c^{2} = 35$ , find the value of $a^{3} + b^{3} + c^{3} - 3 a b c .$

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Solution

a + b + c = 9

Squaring, we get

$(a + b + c)^{2} = (9)^{2} \Rightarrow a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 81 \Rightarrow 35 + 2 (a b + b c + c a) = 81 2 (a b + b c + c a) = 81 - 35 = 46 ∴ a b + b c + c a = \frac{46}{2} = 23 N o w, a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) [a^{2} + b^{2} + c^{2} - (a b + b c + c a)] = 9 [35 - 23] = 9 \times 12 = 108$

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