Question

If a+b+c = 0, then prove that a³+b³+c³=3abc
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Answer 1

Answer:

please refer to the attached picture

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Answer 2

Answer:

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$a + b + c = 0 \\ {a}^{3} + {b}^{3} + {c}^{3} = 3abc \\ (a + b + c) {}^{2} = a {}^{2} + b {}^{2} + c {}^{2} = 2ab + bc + ca) \\ 0 = a {}^{2} + b {}^{2} + c {}^{2} = 2(ab + bc + ca) \\ a {}^{2} + b {}^{2} + c {}^{2} = - 2(ab + bc + ca) \\ a {}^{3} + b {}^{3} + c {}^{3} =( a + b + c)(a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) \\ = 0(same) \\ a {}^{3} + b {}^{3} + c {}^{3} + 3 abc = 0 \\ a {}^{3} + b {}^{3} + c {}^{3} =3 abc(hence \: proved)$