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Find $P(x,y,z)=x^n+y^n+z^n-\prod\limits_{k=0}^{n-1}(x+\omega_n^ky+\omega_n^{-k}z)$, where $\omega_n$ denotes a primitive $n$th root of unity.

I have manually multiplied the terms of the product and then equate the coefficients to get the polynomial but that's too cumbersome. Here is my method : $(x + y + z)(x + y\omega_n + z\omega_n^{n-1})(x + y\omega_n^2 + z\omega_n^{n-2})....(x + y\omega_n^{n-1} + z\omega_n) = x^n(1 + [Y + Z])(1 + [Y\omega_n + Z\omega_n^{n-1}])(1 + [Y\omega_n^2 + Z\omega_n^{n-2}])....(1 + [Y\omega_n^{n-1} + Z\omega_n])$ where $Y=\frac {y}{x}$ and $Z=\frac {z}{x}$ Hence, I applied the formula: $(1+\alpha)(1+\beta)(1+\gamma)...... = 1 + [\alpha + \beta + \gamma + ...] + [\alpha\beta + \beta\gamma + ....] + ....$ to get an expression for odd values of $n$ : $$P = nxyz(x^{n−3}+x^{n−5}yz+x^{n−7}y^2z^2+....)$$ My question is : How can I get a general expression for $P$ in a way better than what I have mentioned? $$EDIT$$ I have a more general expression by now, which I think can be derived elementarily (unfortunately I still don't know how), $$P=\frac {x^n}{t^n}(L_n(t)-t^n),$$ where $L_n(t)$ is the $n^{th}$ Lucas polynomial in $t:=\frac {ix}{\sqrt {yz}}$

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    $\begingroup$ Welcome to math stack exchange. It would be more possible to get answers from others if you show what you have tried, what your thoughts are, etc. More contexts can help others know your situation. Also, it would be better to avoid imperative, as some people get upset about being obliged to answer, say. :-) $\endgroup$
    – awllower
    Commented Oct 20, 2018 at 7:54
  • $\begingroup$ I don't see how it is imperative. $\endgroup$ Commented Dec 8, 2018 at 3:37

1 Answer 1

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We need to find a closed form expression of $$ \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z), $$ where $\omega=e^{\frac{2\pi i}{n}}$ is the primitive $n$-th root of unity. Now let us consider the polynomial $$ F(r) = r^n\prod_{k=0}^{n-1}(x+r\omega^ky+r^{-1}\omega^{-k}z)=\prod_{k=0}^{n-1}(xr+r^2\omega^ky+\omega^{-k}z) $$ of degree $2n$ in $r$ regarding $x,y,z$ as fixed constants. By solving $$ x+r\omega^ky+r^{-1}\omega^{-k}z = 0 $$ for each $0\le k\le n-1$, we can see that $F(r)$ has simple roots $$ r\omega^k = \frac{-x \pm \sqrt{x^2 - 4yz}}{2y}\Longrightarrow r = \omega^{-k}\frac{-x \pm \sqrt{x^2 - 4yz}}{2y}. $$ If we denote $$\alpha(x,y,z) = \frac{-x + \sqrt{x^2 - 4yz}}{2y},\qquad \beta(x,y,z)=\frac{-x - \sqrt{x^2 - 4yz}}{2y},$$ we see that the set of all roots of $F(r)$ coincides with that of $$G(r)=(r^n-\alpha^n(x,y,z))(r^n-\beta^n(x,y,z)).$$ Since neither $F$ nor $G$ has multiple roots, it follows that $F$ is a constant multiple of $G$, i.e. $$ F(r) = \left[\prod_{0\leq k\leq n-1} \omega^ky \right]G(r)=(-1)^{n-1}y^nG(r). $$ by matching the leading coefficient. Plugging $r=1$ into the expression gives \begin{align*} \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z)=&F(1)\\ =& (-1)^{n-1}y^nG(1)\\ =& (-1)^{n-1}y^n(1-\alpha^n-\beta^n +\alpha^n\beta^n). \end{align*}Using $\alpha +\beta = -\tfrac x y$ and $\alpha\beta = \tfrac z y$, we get $$ (-1)^{n-1}y^n(1-\alpha^n-\beta^n +\alpha^n\beta^n)=(-1)^{n-1}(y^n-(y\alpha)^n-(y\beta)^n+z^n ). $$It remains to evaluate $$ (y\alpha)^n+(y\beta)^n = \left(\frac{-x + \sqrt{x^2 - 4yz}}{2}\right)^n + \left(\frac{-x - \sqrt{x^2 - 4yz}}{2}\right)^n. $$ If we let $t= \frac{ix}{\sqrt{yz}}$, we have \begin{align*} (y\alpha)^n+(y\beta)^n =& (i\sqrt{yz})^n\left(\left(\frac{t + \sqrt{t^2 +4}}{2}\right)^n + \left(\frac{t - \sqrt{t^2 +4}}{2}\right)^n\right) \\ =& (-1)^nx^nt^{-n}L_n(t) \end{align*} where $L_n(t)$ is $n$-th Lucas polynomial. Finally, this gives $$ \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z) = (-1)^{n-1}y^n+ (-1)^{n-1}z^n+x^nt^{-n}L_n(t), $$ hence $$ P(x,y,z) = x^n(1-t^{-n}L_n(t)) + (1+(-1)^n)y^n + (1+(-1)^n)z^n. $$

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