## Compute a quotient in a given extension of QQ

Let $\omega$ be a primitive cube root of 1. (That is, $\omega^3 = 1$, but no smaller power of $\omega$ is 1.) Find $a,b,c \in \mathbb{Q}$ such that $2 - \omega = (a+b\omega + c\omega^2)(3 + \omega^2)$.

Carrying out the multiplication on the right hand side and noting that $\omega^3 = 1$, we get the following system of equations: $a+3c = 0$, $3b+c = -1$, and $3a+b = 2$. Solving this system yields $a = 3/4$, $b = -1/4$, and $c = -1/4$. Indeed, we can easily verify that $(\omega^2+3)(\frac{3}{4} - \frac{1}{4}\omega - \frac{1}{4}\omega^2) = 2 - \omega$.