58th Ukrainian National Mathematical Olympiad

Let's add up all three equations:

$$\begin{gathered} (x^3-3x^2+3x)+(y^3-6y^2+12y)+(z^3-9z^2+27z)=189\iff \\ (x^3-3x^2+3x-1)+(y^3-6y^2+3y-27)+(z^3-9z^2+27z-27)=153\iff \\ (x-1{)}^3+(y-2{)}^3+(z-3{)}^3=153. \end{gathered}$$

By a simple exhaustive search, we can see that number 153 has a unique representation in the form of the sum of three cubes: $153=5^3+3^3+1^3$. Therefore, in the last equation, values of $(x-1)$, $(y-2)$ and $(z-3)$ must be selected from 1; 3; 5. Again, simply by trying out the variants, we see that no solution is possible.