If a + b + c = 9 and ab + bc + ca = 26, find the value of $a^{3} + b^{3} + c^{3} - 3 a b c$ .

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Solution

a + b + c = 9, ab + bc + ca = 26

Squaring, we get

$(a + b + c)^{2} = (9)^{2} a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 2 \times 26 = 81 \Rightarrow a^{2} + b^{2} + c^{2} + 52 = 81 ∴ a^{2} + b^{2} + c^{2} = 81 - 52 = 29 N o w, a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) [(a^{2} + b^{2} + c^{2} - (a b + b c + c a))] = 9 [29 - 26] = 9 \times 3 = 27$

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