Problem 3. Factorize

$$
M(x ; y ; z)=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y) .
$$
Solution. Let's assign some values to the variables $x$ and $y$. Let $x=y=1$. Then $M(1 ; 1 ; z)=$ $=(1-z)+(z-1)=0$. Let $x=y=2$. Then $M(2,2, z)=8(2-z)+8(z-2)=0$. This suggests that $M(x, y, z)$ is divisible by $(x-y)$ or $(y-x)$. Thus, we come to the hypothesis that $M(x, y, z)$ is divisible by $(x-z)$ and $(y-z)$.

If $z=0$, then $M(x, y, 0)=x^{3} y-y^{3} x=x y(x-$ $-y)(x+y)$.

If $y=0$, then $M(x, 0, z)=-x^{3} z+z^{3} x=z x(z-$ $-x)(z+x)$.

If $x=0$, then $M(0, y, z)=y^{3} z-z^{3} y=y z(y-$ $-z)(y+z)$.

Thus, we get the hypothesis that

$M(x, y, z)=(x+y+z)(x-y)(y-z)(z-x)$.

It is not difficult to verify its correctness.