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$$(x + y + z)(x + y\omega_n + z\omega_n^{n-1})(x + y\omega_n^2 + z\omega_n^{n-2})....(x + y\omega_n^{n-1} + z\omega_n) = x^n + y^n + z^n - P(x,y,z)$$ where $\omega_n$ is an nth root of unity. The question is to find the polynomial $P$.

I have tried to manually multiply the terms of LHS and then equate the coefficients to get the polynomial but that's too cumbersome: $(x + y + z)(x + y\omega_n + z\omega_n^{n-1})(x + y\omega_n^2 + z\omega_n^{n-2})....(x + y\omega_n^{n-1} + z\omega_n) = x^n(1 + [Y + Z])(1 + [Y\omega_n + Z\omega_n^{n-1}])(1 + [Y\omega_n^2 + Z\omega_n^{n-2}])....(1 + [Y\omega_n^{n-1} + Z\omega_n])$ where $Y=\frac {y}{x}$ and $Z=\frac {z}{x}$ Hence, we can apply the formula: $(1+\alpha)(1+\beta)(1+\gamma)...... = 1 + [\alpha + \beta + \gamma + ...] + [\alpha\beta + \beta\gamma + ....] + ....$ I hope someone can help.

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    $\begingroup$ Isn't that A113279? $\endgroup$ – Bullet51 Oct 21 '18 at 10:24
  • $\begingroup$ If $n$ is odd and $\omega_n$ is a primitive $n$-th rooth of unity, then it is immediate to check that $P(x,y,0)=P(x,0,z)=P(0,y,z)=0$, so $P(x,y,z)=xyz \cdot Q(x,y,z)$ for some polynomial $Q(x,y,z)$ in this case. $\endgroup$ – Maurizio Moreschi Oct 21 '18 at 19:18
  • $\begingroup$ For odd $n$ I have found the following expression: $$P = nxyz(x^{n-3} + x^{n-5}yz + x^{n-7}y^2z^2 + ....)$$ $\endgroup$ – Awe Kumar Jha Oct 22 '18 at 3:34
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As suggested in the first comment, computationally it looks like $$P\equiv P_n=\dfrac{x^n}{t^n}L_n(t)-x^n=\dfrac{x^n}{t^n}(L_n(t)-t^n), $$

where $L_n$ is the $n$th Lucas polynomial in $t:=\dfrac{{ix}}{\sqrt{yz^{\phantom l}}}$.

E.g. for $n=6$, $$P=6x^4yz-9x^2y^2z^2+2xy^3z^3$$ while $$L_6(t)=t^6+6t^4+9t^2+2.$$

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    $\begingroup$ For every $n$ consider a matrix $M_n=(m_{i,j})$ with entries $m_{i,i}=x$, $m_{i,i+1}=y$, and $m_{i,i-1}=z$ for all $i\in[1,n]$ where indices are read modulo $n$, and $m_{i,j}=0$ otherwise. Then $M_n$ is a circulant matrix and its determinant is equal to $\prod_{k=0}^{n-1} (x+\omega_n^ky+\omega_n^{n-k}z)$. Laplace expansion with respect to the last row yields a recursion for $\operatorname{det}(M_n)$, which we could perhaps use to prove your formula. $\endgroup$ – Philipp Lampe Oct 22 '18 at 17:15

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