I derived a theorem regarding cyclic sums of three variables.
Define $\sigma:\mathbb R^3\to\mathbb R$ by $$\sigma(x,y,z)=xy+yz+zx.$$ Given $\alpha,\beta,\gamma\in\mathbb R$ such that $\alpha+\beta+\gamma = 0$, let $x,y,z\in\mathbb R$ be such that $$\alpha x+\beta y + \gamma z=0.$$ Then also $$\alpha \sigma(-x,y,z)^2+\beta\sigma(x,-y,z)^2+\gamma\sigma(x,y,-z)^2=0.\tag 1$$ for which $\sigma(\cdot)^2=\big(\sigma(\cdot)\big)^2$.
My question is, can it be generalised to cyclic functions (or sums in general) of $n$ variables?
I present a few corollaries to motivate interest on this theorem.
Consider $(\alpha,\beta,\gamma)=(-1,2,-1)$, then we have: if $x=2y-z$, then $$({-xy}+yz-xy)^2=2({-xy}-yz+zx)^2-(xy-yz-zx)^2.$$ Since the equation $x=2y-z$ can be solved in $p,q,r$ as $$(x,y,z)=(p+2q,p+q+r,p+2r),$$ it follows that $$\begin{align}\sigma(-x,y,z)&=-p^2-2q(2p+q+2r)+2r^2 \\ \sigma(x,-y,z)&=-p^2-2p(q+r)-2(q^2+r^2)\\ \sigma(x,y,-z)&=-p^2-2r(2p+2q+r)+2q^2.\end{align}$$ We arrive thusly at the following identity for all $p$, $q$ and $r$. $$\big(p^2+2q(2p+q+2r)-2r^2\big)^2=2\big(p^2+2p(q+r)+2(q^2+r^2)\big)^2-\big(p^2+2r(2p+2q+r)-2q^2\big)^2.$$
Not to mention other corollaries like the following: Given $xyz\neq 0$, if $$\alpha+\beta+\gamma=x+y+z=0,$$ then: $$\alpha\bigg({-\frac\alpha x}+ \frac\beta y + \frac\gamma z\bigg)^2+\beta \bigg(\frac\alpha x -\frac\beta y + \frac\gamma z\bigg)^2+\gamma \bigg(\frac\alpha x + \frac\beta y - \frac\gamma z\bigg)^2=0.\tag 2$$
We can show this by dividing Eq. $(1)$ through $x^2y^2z^2$, substituting $\sigma(a_1,a_2,a_3)\mapsto -\sigma(a_1,a_2, a_3)$ since squares are even functions so this substitution does not change the value of the equation, and then lastly substituting $x\mapsto x/\alpha$, $y\mapsto y/\beta$ and $z\mapsto z/\gamma$ to complete the derivation.
We can then use this corollary to obtain a wild modification of $(1)$. Namely, we let $(x,y,z)=(p,q,r)$ in $(2)$. Then in $(1)$, we substitute $$\begin{align}x&\mapsto \bigg({-\frac{\alpha}p}+\frac{\beta}q+\frac{\gamma}r\bigg)^2\\ y&\mapsto \bigg(\frac{\alpha}p-\frac{\beta}q+\frac{\gamma}r\bigg)^2\\ z&\mapsto \bigg(\frac{\alpha}p+\frac{\beta}q-\frac{\gamma}r\bigg)^2.\end{align}$$ and the theorem would still be true but $\sigma(x,y,z)$ would be an algebraic mess.
I derived $(1)$ experimentally upon noticing that $3\times 5 - 5\times 7 + 7\times 3 = 1$ and was so intrigued by this that I considered $xy+yz+zx$ for consecutive odd integers $x,y,z$ ordered by $x< y<z$. I then composed a table of values and did the same thing for $-xy+yz+zx$, $xy-yz+zx$ and $xy+yz-zx$ after which I compared their values and noticed a pattern. I extended this to the present result of $(1)$ but I do not know how to begin for cyclic equations like $xyz+yzw+zwx+wxy$ and etc.
Thank you in advance. Apologies if the question is too long and/or open-ended.