\left.\begin{array}{l}{[\text { Formulas for abbreviated multiplication (etc.). }]} \\ {[\quad \underline{\text { Divisibility rules for } 3 \text { and 9 }}}\end{array}\right]

Do there exist such integers $x, y$ and $z$ for which the equality holds: $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=$ $2011 ?$
Expanding the brackets and combining like terms, we get: $-3 x^{2} y+3 x y^{2}-3 y^{2} z+3 z y^{2}+3 z^{2} x-3 z x^{2}=2011$. The left side is a multiple of 3, while the right side is not. This is a contradiction.

## Answer

They do not exist.