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?


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.


$p_1= x+y+z = e_1$

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

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

$e_2 = xy + xz + yz$

$e_3 = xyz$


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

  • $\begingroup$ You've just proved Newton's identities... $\endgroup$ – lhf Oct 29 '13 at 11:23
  • 4
    $\begingroup$ @lnf sort of. but for Newton's identities of this low degree, it will be useful to explicitly spell out everything. $\endgroup$ – achille hui Oct 29 '13 at 11:29

\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}


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 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.


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.

  • $\begingroup$ Sorry but I know nothing about integral domains. Could you explain more about the $u(x,y)$ divides constant term part? $\endgroup$ – AaronS Oct 30 '13 at 0:13
  • $\begingroup$ @Aaron San: added notes about integral domains $\endgroup$ – miracle173 Nov 2 '13 at 19:27

$$\underbrace{x^3+y^3}+z^3-3xyz = \underbrace{(x+y)^3-3xy(x+y)}+z^3-3xyz$$

$$=\underbrace{(x+y)^3+z^3}-\underbrace{3xy(x+y)-3xyz} $$ $$=\underbrace{\{(x+y)+z\}}\{(x+y)^2-(x+y)z+z^2\}-3xy\underbrace{\{(x+y)+z\}} \left(\text{ using } a^3+b^3=(a+b)(a^2-ab+b^2)\text{ for the first two terms }\right) =(x+y+z)\{(x+y)^2-(x+y)z+z^2-3xy\}$$

Now $\displaystyle (x+y)^2-(x+y)z+z^2-3xy=x^2+y^2+z^2-xy-yz-zx$

  • $\begingroup$ Aha! Did you accidentally delete your answer before? $\endgroup$ – AaronS Nov 30 '13 at 15:53
  • $\begingroup$ @AaronS, no no. I added it a little back as it was required here(math.stackexchange.com/questions/586906/…) $\endgroup$ – lab bhattacharjee Nov 30 '13 at 15:55
  • $\begingroup$ @AaronS, I request you to not to edit so soon :). Now, I can not keep your formatting as I wanted to add inline explanation $\endgroup$ – lab bhattacharjee Nov 30 '13 at 16:11

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