Answer

Let

a = x-y
b = y-z
c = z-x

Addition of the three equations results in

a+b+c = 0

For sum of three cubes, there is the following identity

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Then,

a^3+b^3+c^3-3abc = 0

Substituting back in terms of x,y,z yields

(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-x)(z-x)

Steven Zheng posted 2 weeks ago

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