Without finding the cubes,factorise $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}$.

(D) We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.
If $a+b+c = 0$,then $a^{3}+b^{3}+c^{3} = 3abc$.
Let $a = (x-y)$,$b = (y-z)$,and $c = (z-x)$.
Now,calculate the sum $a+b+c = (x-y) + (y-z) + (z-x) = x - x + y - y + z - z = 0$.
Since the sum is $0$,we can apply the identity $a^{3}+b^{3}+c^{3} = 3abc$.
Therefore,$(x-y)^{3}+(y-z)^{3}+(z-x)^{3} = 3(x-y)(y-z)(z-x)$.