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If a+b+c=15 and a^2+b^2+c^2=83 then find the value of a^3+b^3+c^3-3abc.

Answer» a + b + c = 15 \: \: \: \: \: and \: \: \: \: {a}^{2} + {b}^{2} + {c}^{2} = 83 \\ (a + b + c) ^{2} = {15}^{2} \\ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 225 \\ 2(ab + bc + ca) = 225 - 83 \\ 2(ab + bc + ca) = 142 \\ ab + bc + ca = 71 \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)[ a^{2} + b ^{2} + c ^{2} - (ab + bc + ca) ] \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc = 15(83 - 71) \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc= 15 \times 12 \\{a}^{3} + {b}^{3} + {c}^{3} - 3abc= 180


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