Let $f(x, y, z)$ be the numerator, i.e. ${(y-z)^3 +(z-x)^3+(x-y)^3}$. This is a symmetric expression. When $x=y$, $f(x,y,z)=0$. Therefore, $(x-y)$ divides $f(x,y,z)$. By symmetry, it must hold that both $(y-z)$ and $(z-x)$ are divisors of $f(x, y, z)$. Hence $$f(x, y, z)=k(x-y)(y-z)(z-x)$$ where $k$ is a constant. Setting $x=0$, $y=1$ and $z=-1$ yields $$f(0, 1, -1)=6\quad\text{and}\quad k(x-y)(y-z)(z-x)=2k$$ Hence, $k=3$, therefore $$\frac{(y-z)^3 +(z-x)^3+(x-y)^3}{(y-z)(z-x)(x-y)}=\frac{3(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}=\boxed{3}$$