<s>[INST] 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?
 [/INST] 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.
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