Factorize the polynomial $x^3+y^3+z^3-3xyz$

\begin{align}x^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \\ & =(x+y)^3+z^3-3xy(x+y+z)\\ &= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \\ & =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \\ & =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\end{align}

Consider the polynomial $$(\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - a\lambda^2+b\lambda-c\tag{*1}$$ We know $$\begin{cases}a = x + y +z\\ b = xy + yz + xz \\ c = x y z\end{cases}$$ Substitute $x, y, z$ for $\lambda$ in $(*1)$ and sum, we get $$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$ This is equivalent to $$\begin{align} x^3+y^3+z^3 - 3xyz = & x^3+y^3+z^3 - 3c\\ = & a(x^2+y^2+z^2) - b(x+y+z)\\ = & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx) \end{align}$$

Note that (can be easily seen with rule of Sarrus)$$ \begin{vmatrix} x & y & z \\ z & x & y \\ y & z & x \\ \end{vmatrix}=x^3+y^3+z^3-3xyz $$

On the other hand, it is equal to (if we add to the first row 2 other rows) $$ \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ y & z & x \\ \end{vmatrix}=(x+y+z)\begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ y & z & x \\ \end{vmatrix}=(x+y+z)(x^2+y^2+z^2-xy-xz-yz) $$ just as we wanted. The last equality follows from the expansion of the determinant by first row.