When $X+Y+Z=0$ $X^3+Y^3+Z^3=3XYZ$

Hence, $$\boxed {3 (a+2b-3c)(b+2c-3a)(c+2a-3b)} $$

We know that $(a+2b-3c)+(b+2c-3a)+(c+2a-3b)=0.$ Thus, $[(a+2b-3c)+(b+2c-3a)+(c+2a-3b)]^3=0.$ Hence, we get the factorization $\boxed{3 (a+2b-3c)(b+2c-3a)(c+2a-3b)}$.