Selection and Training Session

First, note that $a\ne 0$, $x\ne -y$, etc. (this easily follows from the given relation). Now, the equality $x^2-1/y=y^2-1/z$ can be written in two other ways: $$\begin{array}{rl}{x}^2-y^2+\frac{1}{x}-\frac{1}{y}& =\frac{1}{x}-\frac{1}{z}\iff (x-y)(x+y-\frac{1}{xy})=\frac{z-x}{xz}\\ (x-y)(x^2+xy-\frac{1}{y})& =\frac{z-x}{z}\iff (x-y)(a+xy)=\frac{z-x}{z}\end{array}(1)$$ and $$\begin{array}{rl}{x}^2-y^2-\frac{1}{x}-\frac{1}{y}& =\frac{1}{x}-\frac{1}{z}\iff (x+y)(x-y-\frac{1}{xy})=-\frac{z+x}{xz}\\ (x+y)(x^2-xy-\frac{1}{y})& =-\frac{z+x}{z}.\end{array}(2)$$ Now, multiplying (1) by two similar equalities and reducing by $(x-y)(y-z)(z-x)\ne 0$, we obtain $(a+xy)(a+yz)(a+zx)=\frac{1}{xyz}$ $$a^3+a^2(xy+yz+zx)+a(x+y+z)xyz+(xyz{)}^2=\frac{1}{xyz}.(3)$$ Proceeding similarly with (2), we obtain $$a^3-a^2(xy+yz+zx)+a(x+y+z)xyz-(xyz{)}^2=-\frac{1}{xyz}.(4)$$ Summing (3) and (4) and reducing by $2a\ne 0$, we obtain the required equality.