<s>[INST] factorization of $a^3 + b^3$ over $\mathbb{C}$ I have been trying to understand how to factorize a basic $a^3 + b^3$  over $\mathbb{C}$. I saw in the book that it is $ \Pi_{i=1}^{3}(a + \omega^{(2i+1)}b)$ where $\omega$ is the cuberoot of  $-1$.
I was trying to derive it in the following way and was surprised by something which I cannot explain. $a^3 + b^3 = (a^3 - (\omega b)^3)$, I tried to use the formula for factorization of $a^3 - b^3$. Since we know $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ hence I was trying to factorize  the original expression as $a^3 -(\omega b)^3 = (a - \omega b)(a^2 + \omega ab + (\omega b)^2)$.
Then using $ 1 + \omega + \omega^2 = 0$. I substituted $(a -\omega b)(a^2 -(1 + \omega^2) ab + (\omega b)^2) $. This is factored as $(a -\omega b)(a - \omega ^2 b ) (a - b) $. This when expanded is $ (a^2 - \omega^2 ab - \omega ab + \omega^3 b^2) = (a^2 + ab - b^2) (a -b)$. Surely this product does not give back the original expression.
Which assumption am I misunderstanding during substitution which is creating this problem is unclear.
 [/INST] Not exactly sure what our OP user2714795 is trying to do, but here's my approach:
Does
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)? \tag 1$
we compute
$(a + b)(a^2 - ab + b^2)$
$= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3, \tag 2$
so yes, (1) binds.  We also have
$(a + b\omega)(a + b \bar \omega) = a^2 + ab (\omega + \bar \omega) + b^2 \omega \bar \omega \tag 3$
for any
$\omega \in \Bbb C; \tag 4$
if we take
$\omega = -\dfrac{1}{2} + \dfrac{\sqrt 3}{2}i \tag 5$
then (3) becomes
$(a + b\omega)(a + b \bar \omega)$
$= a^2 + ab (\omega + \bar \omega) + b^2 \omega \bar \omega = a^2 - ab + b^2;  \tag 6$
combining (1) and (6) yields
$a^3 + b^3 = (a + b)(a + b\omega)(a + b\bar \omega), \tag 7$
and we have factored $a^3 + b^3$ over $\Bbb C$.
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