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 Oct 20 '18 at 7:54
  • $\begingroup$ I don't see how it is imperative. $\endgroup$ – Awe Kumar Jha Dec 8 '18 at 3:37

The problem comes down to finding a formula for $$ \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z), $$ where $\omega=e^{\frac{2\pi i}{n}}$ is the $n$-th root of unity. Let us consider the expression $$ F(r) = \prod_{k=0}^{n-1}(x+r\omega^ky+r^{-1}\omega^{-k}z). $$ Consider this expression as a function of $r$ and regard $x,y,z$ as constant. Solve the equation $F(r)=0$. Then, it holds that $$ r\omega^k = \frac{-x \pm \sqrt{x^2 - 4yz}}{2y}. $$ Define $\alpha = \alpha(x,y,z) = \frac{-x + \sqrt{x^2 - 4yz}}{2y}$ and $\beta= \beta(x,y,z)=\frac{-x - \sqrt{x^2 - 4yz}}{2y}$. We can see that $F(r)=0$ has every $n$-th root of $\alpha^n$ and $\beta^n$, hence giving us the result $$ F(r) = (\prod_{0\leq k\leq n-1} \omega^ky)\cdot r^{-n}(r^n-\alpha^n)(r^n-\beta^n). $$ Plugging $r=1$ into the expression gives us $$ F(1)= \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z)= (-1)^{n-1}y^n(1-\alpha^n)(1-\beta^n) = (-1)^{n-1}y^n(1-\alpha^n-\beta^n +\alpha^n\beta^n). $$Note that $\alpha +\beta = -x/y$ and $\alpha\beta = z/y$. Hence we get the expression $$ (-1)^{n-1}(y^n+z^n-(y\alpha)^n-(y\beta)^n ). $$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 $$ (i\sqrt{yz})^n\left(\left(\frac{t + \sqrt{t^2 +4}}{2}\right)^n + \left(\frac{t - \sqrt{t^2 +4}}{2}\right)^n\right) = (-x)^nt^{-n}L_n(t). $$ Therefore, we get that $$ \prod_{k=0}^{n-1}(x+\omega^ky+\omega^{-k}z) = (-1)^{n-1}y^n+ (-1)^{n-1}z^n+x^nt^{-n}L_n(t), $$ and $$ P = x^n(1-t^{-n}L_n(t)) + (1+(-1)^n)y^n + (1+(-1)^n)z^n. $$


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