rbs wrote: Let $x,y,z$ be 3 different real numbers not equal to $0$ that satisfiying $x^2-xy=y^2-yz=z^2-zx$. Find all the values of $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}$ and $(x+y+z)^3+9xyz$. Solving the first system and considering $x,y,z$ nonzero and pairwise different, we get the general solution : $(x,y,z)=(u^2v,u^2v+uv-v,uv)$ whatever is $v\ne 0$ and for some $u$ root of $u^3+3u^2-1=0$ (which exists) Then $\frac xz+\frac yx+\frac zy=\frac{u^5+2u^4+2u^3-u^2-2u+1}{u^4+u^3-u^2}$ $=\frac{-3(u^4+u^3-u^2)+(u^3+3u^2-1)(u^2+2u-1)}{u^4+u^3-u^2}=-3$ And so $\boxed{\frac xz+\frac yx+\frac zy=-3}$ And $(x+y+z)^3+9xyz=v^3(8u^6+33u^5+21u^4-25u^3-6u^2+6u-1)$ $=v^3(u^3+3u^2-1)(8u^3+9u^2-6u+1)=0$ And so $\boxed{(x+y+z)^3+9xyz=0}$

rbs wrote: Let $x,y,z$ be 3 different real numbers not equal to $0$ that satisfiying $x^2-xy=y^2-yz=z^2-zx$. Find all the values of $\frac{x}{z}+\frac{y}{x}+\frac{z}{y}$ and $(x+y+z)^3+9xyz$. Solution. Put $x^2-xy=y^2-yz=z^2-zx=m$. Since $x,y,z$ are different nonzero numbers, $m\ne0$, and thus \begin{align*}\begin{cases}x-y=\frac{m}{x}\\ y-z=\frac{m}{y}\\ z-x=\frac{m}{z}\end{cases}\Longrightarrow&\begin{cases}xy-y^2=m\cdot\frac{y}{x}\\ yz-z^2=m \cdot\frac{z}{y}\\ zx-x^2=m\cdot\frac{x}{z}\end{cases}\\ \Longrightarrow& xy-y^2+yz-z^2+zx-x^2=m\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)\\ \Longrightarrow& -3m=m\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)\\ (\because m\ne0)\Longrightarrow& -3=\frac{y}{x}+\frac{z}{y}+\frac{x}{z}.\end{align*}\begin{align*}\begin{cases}x-y=\frac{m}{x}\\ y-z=\frac{m}{y}\\ z-x=\frac{m}{z}\end{cases}\Longrightarrow&\frac{m}{x}+\frac{m}{y}+\frac{m}{z}=0=\frac{mz}{x}+\frac{mx}{y}+\frac{my}{z}\\ (\because m\ne0)\Longrightarrow&\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=\frac{z}{x}+\frac{x}{y}+\frac{y}{z}\\ (\text{multiply by }xyz)\Longrightarrow&yz+xz+xy=0=z^2y+x^2z+y^2x\\ \Longrightarrow&(x+y+z)^2=x^2+y^2+z^2=3m\\ \Longrightarrow&(x+y+z)^3=3m(x+y+z), \end{align*}while \begin{align*}m(x+y+z)=&mx+my+mz\\ =&\left(y^2-yz\right)x+\left(z^2-xz\right)y+\left(x^2-xy\right)z\\ =&y^2x+z^2y+x^2z-3xyz\\ =&-3xyz.\end{align*}We are done. $\blacksquare$