<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] 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).
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