Question

Q22.If a + b + c = 6, a^2 + b^2 + c^2 = 14 and a^3 + b^3 + c^3 = 36, find the value of abc.

Answer 1

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$we \: know \: \\ ( {a + b + c})^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) \\ (6) ^{2} = 14 + 2(ab + bc + ca) \\ 36 - 14 = 2(ab + bc + ca) \\ ab + bc + ca = \frac{22}{2} = 11 \\ now \\ {a}^{3} + {b}^{3} + {c}^{3} - 3bc \\ = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ 36 - 3abc = 6(14 - 11) \\ 36 - 3abc = 18 \\ 3abc = 18 \\ abc = 6$