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Let $\zeta_p$ be a $p$-th root of unity, where $p$ is an odd prime number.

I just came across the following expression:

$$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2}.$$ Can we simplify this expression somehow? For $p=3$, I can rewrite this to $$\frac{8}{(\zeta_3-1)^2}.$$

Question. Can we simplify $$\frac{(\zeta_p^2-\zeta_p+1)^3}{\zeta_p^2(\zeta_p-1)^2} ?$$

Note that this expression does not change if we replace $\zeta_p$ by one of the elements in $\{\zeta_p, \zeta_p^{-1},1-\zeta_p, (1-\zeta_p)^{-1}, \zeta_p/(1-\zeta_p), (1-\zeta_p)/\zeta_p\}$.

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    Honestly, the expression looks simple to me, particularly for large $p$.2011-10-14
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    The reason that you are successful in simplifying when $p=3$ is that the top factors as $$ (\zeta_p-\zeta_6)(\zeta_p-\zeta_6^5) $$ where $\zeta_6$ is a primitive sixth root of unity; since $\zeta_6$ is $\zeta_3 + 1$ (for an appropriate choice of $\zeta_3$), you can simplify. I agree with Swapan that there doesn't seem to be any general reason why the expression would simplify for larger $p$.2011-10-14

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