jmc

Combining all three fractions under a single denominator, the given expression is equal to $$\frac{(x-y{)}^3+(y-z{)}^3+(z-x{)}^3}{(x-y)(y-z)(z-x)}.$$ Consider the numerator as a polynomial in $x$, so that $P(x)=(x-y{)}^3+(y-z{)}^3+(z-x{)}^3$ (where we treat $y$ and $z$ as fixed values). It follows that $P(y)=(y-y{)}^3+(y-z{)}^3+(z-y{)}^3=0$, so $y$ is a root of $P(x)=0$ and $x-y$ divides into $P(x)$. By symmetry, it follows that $y-z$ and $z-x$ divide into $P(x)$. Since $P$ is a cubic in its variables, it follows that $P=k(x-y)(y-z)(z-x)$, where $k$ is a constant. By either expanding the definition of $P$, or by trying test values (if we take $x=0,y=-1,z=1$, we obtain $P=-6=k\cdot (-2)$), it follows that $k=3$. Thus, $$\frac{(x-y{)}^3+(y-z{)}^3+(z-x{)}^3}{(x-y)(y-z)(z-x)}=\boxed{3}.$$