Factorize the polynomial $x^3+y^3+z^3-3xyz$ I want to factorize the polynomial $x^3+y^3+z^3-3xyz$. Using Mathematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?
 A: $$\underbrace{x^3+y^3}+z^3-3xyz = \underbrace{(x+y)^3-3xy(x+y)}+z^3-3xyz$$
$$=\underbrace{(x+y)^3+z^3}-\underbrace{3xy(x+y)-3xyz} $$
$$=\underbrace{\{(x+y)+z\}}\{(x+y)^2-(x+y)z+z^2\}-3xy\underbrace{\{(x+y)+z\}} \left(\text{ using } a^3+b^3=(a+b)(a^2-ab+b^2)\text{ for the first two terms }\right)
=(x+y+z)\{(x+y)^2-(x+y)z+z^2-3xy\}$$
Now $\displaystyle (x+y)^2-(x+y)z+z^2-3xy=x^2+y^2+z^2-xy-yz-zx$
A: 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.
A: \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}
A: 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
$$\begin{align}
  x^3+y^3+z^3 - 3xyz
= & x^3+y^3+z^3 - 3c\\
= & a(x^2+y^2+z^2) - b(x+y+z)\\
= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)
\end{align}$$
A: 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: 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.
A: I usually work with students on this classic problem as follows:

*

*Show that if $x = -y-z$ the polynomial vanishes. Hence $x-(-y-z)$ is a factor.  This is an application of a general method that is often used for factoring polynomials in one variable, but not in more than one.  It also helps students see that a polynomial in several variables can be seen as defining a function of any one (or several) of these variables, with the others held constant.

*The polynomial is (a) homogeneous and (b) symmetric in $x$, $y$, $z$. Good review of those properties.

*So it factors as $(x+y+z)(Ax^2 + Ay^2 + Az^2 + Bxy + Byz + Bxz)$.  Then let $x = 0$ and we get $(y+z)(\text{something}) = y^3 + z^3$.  This is a well-known factorization, and the students can easily tell you what 'something' is.  Then we quickly get $A = 1$, $B = -1$.

Then we often go on to derive Cardano's (Tartaglia's!) cubic formula from this factorization.
