Prove that :(a-b)^3+(b-c)^3+(c-a)^3 = 3(a-b) (b-c) (c-a)

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a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} =

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

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

a + b + c = 0

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

> 0

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

\frac{(a + b)^{3} + (b + c)^{3} + (c + a)^{3} - 3 (a + b) (b + c) (c + a)}{3 (a^{3} + b^{3} + c^{3} - 3 a b c)}

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