Question

Eliminate $x,y,z$ from the equations $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=a,\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=b,(\frac{x}{y}+\frac{y}{z})(\frac{y}{z}+\frac{z}{x})(\frac{z}{x}+\frac{x}{y})=c$.

Answer 1

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=a\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=bab=(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})(\frac{x}{z}+\frac{y}{x}+\frac{z}{y})ab=\frac{x^2}{yz}+1+\frac{xz}{y^2}+\frac{yx}{z^2}+\frac{y^2}{zx}+1+1+\frac{zy}{x^2}+\frac{z^2}{xy}ab=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}+\frac{zy}{x^2}+\frac{xz}{y^2}+\frac{yx}{z^2}+3ab=\sum \frac{x^2}{yz}+\sum \frac{z^2}{xy}+\mathrm{3........}\left(i\right)$

$(\frac{x}{y}+\frac{y}{z})(\frac{y}{z}+\frac{z}{x})(\frac{z}{x}+\frac{x}{y})=c(\frac{x}{z}+\frac{z}{y}+\frac{y^2}{z^2}+\frac{y}{x})(\frac{z}{x}+\frac{x}{y})=c1+\frac{x^2}{zy}+\frac{z^2}{xy}+\frac{zx}{y^2}+\frac{y^2}{zx}+\frac{xy}{z^2}+\frac{yz}{x^2}+1=c\frac{x^2}{zy}+\frac{y^2}{zx}+\frac{z^2}{xy}+\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}+2=c\sum \frac{x^2}{zy}+\sum \frac{yz}{x^2}+2=c\sum \frac{x^2}{zy}+\sum \frac{yz}{x^2}=c-2$

substituting in $\left(i\right)$

$\Rightarrow ab=c-2+3\Rightarrow ab=c+1$