Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is 
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
 A: we have
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$$$=a^2+ab\Omega^2+ac\Omega+ab\Omega+b^2\Omega^3+bc\Omega^2+ac\Omega^2+bc\Omega^4+c^2\Omega^3$$
$$=a^2+\Omega^3b^2+\Omega^3c^2+(\Omega^2+\Omega)ab+(\Omega^2+\Omega^4)bc+(\Omega+\Omega^2)ac$$
$$=a^2+b^2+c^2-ab-bc-ac$$
It follows that $$\Omega^3=1$$
and
$$\Omega^2+\Omega=\Omega^2+\Omega^4=-1.$$
Solving $$\Omega^2+\Omega=-1$$ will give you two solutions that work (check that they satisfy all of the above equations).  These happen to be the primitive third roots of unity as $$\Omega^2+\Omega+1$$ is the third cyclotomic polynomial.
A: This is equal to:
$$\frac{a^3+b^3+c^3-3abc}{a+b+c}$$
which you can verify by multiplying your polynomial by $a+b+c$. So let's try to factor $a^3+b^3+c^3-3abc$.
So we know that
$$a+b+c\mid a^3+b^3+c^3-3abc$$
(The vertical line means "is a factor of.")
Now, let $\Omega$ be a cube root of unity (usually denoted by $\omega$ but i'll follow the notation in your question). Notice that the maps:
\begin{align}
a&\mapsto a\\
b&\mapsto b\Omega\\
c&\mapsto c\Omega^2\\
\end{align}
don't change $a^3+b^3+c^3-3abc$. Thus, applying them to the previous equation, we get:
\begin{align}
a+b+c&\mid a^3+b^3+c^3-3abc\\
a+b\Omega+c\Omega^2&\mid a^3+b^3+c^3-3abc\\
a+b\Omega^2+c\Omega&\mid a^3+b^3+c^3-3abc
\end{align}
These are its only factors, since it's a monic degree-3 polynomial and we've just supplied three monic degree-1 factors. Thus:
$$(a+b+c)(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)\\=a^3+b^3+c^3-3abc$$
Dividing out by $a+b+c$ and using the observation at the start of this answer, we get your factorization.
A: If we are given that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = (a + b\Omega + c\Omega^2)(a + b\Omega^2 + c\Omega), \tag{1}$
and we wish to find $\Omega$, we can proceed as follows: multiplying out the right-hand side, we find
that
$a^2 + b^2 + c^2 - ab - ac - bc$
$ = a^2 + \Omega^3 b^2 + \Omega^3 c^2 
$
$+ (\Omega^2 + \Omega)ab + (\Omega^2 + \Omega)ac + (\Omega^4 + \Omega^2) bc; \tag{2}$
we see from (2) that the factorization succeeds provided that
$\Omega^3 = 1, \tag{3}$
$\Omega^2 + \Omega = -1, \tag{4}$
and
$\Omega^4 + \Omega^2 = -1; \tag{5}$
it follows from (3) that $\Omega$ must be one of the three cube roots of unity $1$, $e^{2\pi /3}$, $e^{4\pi i /3}$; it follows from (4), (5) that $\Omega \ne 1$; thus, since
$(\Omega - 1)(\Omega^2 + \Omega + 1) = \Omega^3 - 1 = 0, \tag{6}$
(4) must bind; the apparent discrepancy 'twixt (4) and (5) is resolved by observing that (3) yields
$\Omega^4 = \Omega \Omega^3 = \Omega(1) = \Omega; \tag{7}$
(4) and (5) are in reality the same equation.  We thus see that the factorization (1) holds with $\Omega = e^{2 \pi i / 3}$, $\Omega = e^{4 \pi i / 3}$, since both satisfy (3)-(5).
A: Let's do it as follows for a change.
Consider the circulant matrix
$$
M(a,b,c):=\left(\begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array}\right)
=a I+ b T+ c T^2,
$$
where $T=M(0,1,0)$ satisfies the relations $T^2=M(0,0,1), T^3=M(1,0,0)=I$. 
Let's calculate $\det M(a,b,c)$. Let's first add the 2nd and 3rd rows to the first. The first row then becomes a constant $a+b+c$. Therefore
$$
\det M(a,b,c)=(a+b+c)\left\vert\begin{array}{ccc}1&1&1\\c&a&b\\b&c&a\end{array}\right\vert=(a+b+c)\left[(a^2-bc)+(b^2-ac)+(c^2-ab)\right],
$$
IOW apart from the factor $a+b+c$ we see that your polynomial is the determinant of $M(a,b,c)$.
But the eigenvalues of $T$ are $\lambda_1=1$, $\lambda_2=\omega:=e^{2\pi/3}$ and $\lambda_3=\omega^2$ - the third roots of unity. Therefore the eigenvalues of $M(a,b,c)$ are $a\lambda_i^0+b\lambda_i^1+c\lambda_i^2, i=1,2,3.$ These evaluate to 
$$
a+b+c,\quad a+\omega b+\omega^2c,\quad a+\omega^2 b+\omega c.
$$
The determinant of a matrix is, of course, the product of its eigenvalues. Cancelling the known extra factor $a+b+c$ gives you what you want.

This is, of course, overkill for the purposes of factoring this quadratic, but 
the method generalizes (once you recognize what is required), and circulant matrices are nice to chat about.
