Problem 4. Factorize $M(x, y, z)=$ $=x^{3}+y^{3}+z^{3}-3 x y z$.
Solution. If $x=y=1$, then

$$
\begin{gathered}
M(1,1, z)=2+z^{3}-3 z=\left(z^{3}-z\right)+(2-2 z)= \\
=(z-1)^{2}(z+2)
\end{gathered}
$$

If $x=1, z=2$, then $M(1, y, 2)=y^{3}-6 y+9$. It is easy to notice that the number -3 is a root of the equation $y^{3}-6 y+9=0$. Therefore, by the theorem of Bezout, we get:

$$
y^{3}-6 y+9=(y+3)\left(y^{2}-3 y+3\right)
$$

From equalities (1) and (2), it is not difficult to assume that $M(x, y, z)$ is divisible by $(x+y+z)$. Indeed,

$$
\begin{aligned}
& \begin{array}{l}
\quad x^{3}+y^{3}+z^{3}-3 x y z \\
-x^{3}+x^{2} y+x^{2} z
\end{array}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-049.jpg?height=115&width=742&top_left_y=2187&top_left_x=974)

$$
\begin{aligned}
& -x^{2}(y+z)-3 x y z+y^{3}+z^{3} \\
& -x^{2}(y+z)-x\left(y^{2}+y z\right)-x\left(y z+z^{2}\right) \\
& \begin{array}{l}
x\left(y^{2}+z^{2}-y z\right)+y^{3}+z^{3} \\
x\left(y^{2}+z^{2}-y z\right)+y\left(y^{2}+z^{2}-y z\right)+z\left(y^{2}+z^{2}-y z\right)
\end{array} \\
& 0
\end{aligned}
$$

Thus, $M(x, y, z)=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-\right.$ $-z x)$.