If a+b+c is not equal to 0,then prove that a^3+b^{3 }+c^3=3abc,only when a=b=c

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a + b + c = 0

a^{3} + b^{3} + c^{3} = 3 a b c

a + b + c = 5

a b + b c + c a = 10

a^{3} + b^{3} + c^{3} - 3 a b c = - 25

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

a^{3} + b^{3} + c^{3} - 3 a b c

a + b + c = 0

If a+b+c=0 then $(a^{3} + b^{3} + c^{3})$ is
(a) 0 (b) abc (c) 2abc (d) 3abc

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