If $α,β,γ$ belong to ${1,w,w^2}$ then how is number of triplets satisfying $|(aα+bβ+cγ)/(aβ+bγ+cα)| = 1$ equal to $9$?

Question

My book gives this question, where $1$, $w$ and $w^2$ are cube roots of unity. In the solution given it is written that there are 9 ways in total, when $α,β,γ$ are different (6 cases) and when they are the same (3 cases). Now I understood the three cases when $α=β=γ=1$,$w$ or $w^2$, but how do I find the 6 cases when $α,β,γ$ are different? Also, what about the cases when 2 of them are same while the third is different (eg. $α=w,β=1,γ=1$)? How do I tackle this problem?

Answer 1

This should work out although it is a rather brute force method. Assuming $A,b,c\in R$$$$$ Hint 1: $\left|\dfrac{a\alpha+b\beta+c\gamma}{\alpha\beta+\beta\gamma+\alpha\gamma}\right|=1\Rightarrow |a\alpha+b\beta+c\gamma|=|\alpha\beta+\beta\gamma+\alpha\gamma|$

Hint 2: $|z|^2=z\bar z$ Hence, $(a\alpha+b\beta+c\gamma)(a\bar\alpha+b\bar\beta+c\bar\gamma)=(\alpha\beta+\beta\gamma+\alpha\gamma)(\bar\alpha\bar\beta+\bar\beta\bar\gamma+\bar\alpha\bar\gamma)$

Hint 3: On expanding you will get terms involving $|\alpha|^2,|\beta|^2,|\gamma|^2$. These are each respectively equal to $1$, since $|1|=|\omega|=|\omega^2|=1$.

Note: I would have posted this as a comment, but it was too long for a comment.