Đặt \(x-y=c;y-z=a;z-x=b\) thì \(a+b+c=0\Rightarrow c=-\left(a+b\right)\). Ta có
\(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3=a^3+b^3+c^3\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c^3\)
\(=-c\left(a^2-ab+b^2\right)+c\left(a+b\right)^2\)
\(=c\left(-a^2+ab-b^2+a^2+2ab+b^2\right)\)
\(=3abc=3\left(x-y\right)\left(y-z\right)\left(z-x\right)\)