Without finding the cubes,factorise $(x-2 y)^{3}+(2 y-3 z)^{3}+(3 z-x)^{3}$.

(D) Let $a = x-2y$,$b = 2y-3z$,and $c = 3z-x$.
Then,$a+b+c = (x-2y) + (2y-3z) + (3z-x) = 0$.
We know that if $a+b+c = 0$,then $a^3 + b^3 + c^3 = 3abc$.
Substituting the values of $a, b,$ and $c$ back into the identity,we get:
$(x-2y)^3 + (2y-3z)^3 + (3z-x)^3 = 3(x-2y)(2y-3z)(3z-x)$.