<s>[INST] $x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$) Let $x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$). Can someone give me a hint?
Approach:
$x= a^3 + 3a^2 b\omega + 3ab^2\omega^2 + b^3$ and $\omega^2 = -\omega -1$. But then, I'm stuck.
Thanks in advance.
 [/INST] Hint:
$$ (a + b \omega)^3 = (a \omega + b \omega^2)^3 = (a \omega^2 + b)^3.$$
Now write $$\omega = - \frac 1 2 + \frac{\sqrt{-3}}2, \ \ \ \ \  \ \ \ \omega^2 = - \frac 1 2 - \frac{\sqrt{-3}}2,$$
and separate into cases depending on whether $a$ and $b$ are odd or even...
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