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-yz-xz-3xy)\\ &= (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.

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

Use Newton's identities:

$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ and so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.

Here

$p_1= x+y+z = e_1$

$p_2= x^2+y^2+z^2$

$p_3= x^3+y^3+z^3$

$e_2 = xy + xz + yz$

$e_3 = xyz$

A polynomial from $\mathbb{Q}[x,y,z]$ is a polynomial from $\mathbb{Q}[x,y][z]$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $\mathbb{Q}[x,y]$. $$p(z)=z^3-3xy \cdot z +x^3+y^3$$

So we can try our methods to factor a polynomial of degree 3 over an integral domain: If it can be factored then there is a factor of degree $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the constant term of $p(z)$ which is $x^3+y^3$. The latter is can be factored to $(x+y)(x^2-xy+y^2)$ We check each of the possible values $(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)$ for $u(x,y)$ and find that only $p(-x-y)=0$. So $z-(-x-y)$ is a factor.

Note:

One can use Kronecker's method

• to reduce the factorization of a polynomial of $\mathbb{Q}[x,y,z]$ to factoring polynomials in $\mathbb{Q}[x,y]$,
• to reduce the factorization of polynomial of $\mathbb{Q}[x,y]$ to factoring polynomials in $\mathbb{Q}[x,]$
• to reduce the factorization of polynomial of $\mathbb{Q}[x,]$ to factoring numbers in $\mathbb{Z}$

This factoring is possible in a finite number of steps but the number of steps may become to high for practical purpose.

An integral domain is a commutative ring with $1$, where the following holds: $$a \ne 0 \land b \ne 0 \implies ab \ne 0$$ For polynomials $f$, $g$, $h$ $\in I[z]$ this guarantees: $$f=g \cdot h \implies \text{degree}(f)=\text{degree}(g) + \text{degree}(h) \tag{1}$$ compare this to $\mathbb{Z}_4$ which is no integral domain and $(2z^2+1)^2 \equiv 1$ and so $(2z^n+1) \mid 1$. So the polynomial $1$ of degree $0$ has infinitely many divisor. If $I$ is an integral domain $(1)$ guarantees that $z^3+az^2+bz+c \in I[z]$ has a linear factor and therefore zero in $I$ if it is not irreduzible.