Let $\zeta = - \frac{1}{2}+i \frac{\sqrt{3}}{2}$.
I'm trying to solve the following overdetermined systems of equations:
$$ \begin{align} &x_{1}=u+v \\ &x_{2} = \zeta u + \zeta^{2} v \\ &x_{3} = \zeta^{2} u + \zeta v \end{align}$$
Since $\zeta$ is a third root of unity, $\zeta^{3} =1 $.
I also know that $1+\zeta+\zeta^{2} =0$.
I assume we can use these facts to solve the system, but I'm not sure how.
A solution (which is supposedly unique) is $$u= \frac{1}{3} \left( x_{1} + \zeta^{2} x_{2} + \zeta x_{3} \right), \ v= \frac{1}{3} \left( x_{1} + \zeta x_{2} + \zeta^{2} x_{3} \right),$$ but I would prefer not to work backwards.
EDIT: In retrospect, I should have known what to do by just looking at the solution.
To get the expression for $u$, you multiply each of the three expressions for the $x_{i}$ by suitable powers of $\zeta$, and then sum the three expressions together, so that the overall coefficient for $v$ is $1 + \zeta + \zeta^{2} = 0$, and check that the coefficient of $u$ is nonzero.
(Or, alternatively, multiply by suitable powers of $\zeta$ so that the coefficient of $u$ in each equation becomes $1$, so that summing the three equations together you get a coefficient $3$ for $u$, and check that the coefficient for $v$ vanishes.)
This leaves you with only $u$ in terms of the $x_{i}$. Ditto with the roles of $u$ and $v$ exchanged.
Yes, this shows that the solution is unique.