Eliminate $x,y,z$ from equations Eliminate $x,y,z$ from equations
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=a$$ 
 $$\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=b$$
 $$\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right) =c$$
I try this problem with this method.
$$ab=\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)$$
$\implies$$$ab=3+\sum_{cyclic}\frac{x^2}{yz}+\sum_{cyclic}\frac{xy}{z^2}$$
$$c=\left(\frac{x}{y}+\frac{y}{z}\right)\left(\frac{y}{z}+\frac{z}{x}\right)\left(\frac{z}{x}+\frac{x}{y}\right)=2+\sum_{cyclic}\frac{xy}{z^2}+\sum_{cyclic}\frac{x^2}{yz} $$
from above two equations I get,
$ab-3=c-2$
$ab=c+1$.
Is it correct ?or not correct ?
is there exists another way to do it? i want to know solution 

Following is an alternate way to derive $c = ab - 1$.
Let $\lambda_1 = \frac{x}{y}$, $\lambda_2 = \frac{y}{z}$ and $\lambda_3 = \frac{z}{x}$. It is trivial to see $\lambda_1\lambda_2\lambda_3 = 1$.
In terms of $\lambda_1,\lambda_2,\lambda_3$, we have
$$\begin{align}
a &= \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = \lambda_1 + \lambda_2 + \lambda_3\\
b &= \frac{y}{x} + \frac{z}{y} + \frac{x}{z} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3} = \lambda_2\lambda_3 + \lambda_3\lambda_1 + \lambda_1\lambda_2
\end{align}$$
Using Vieta's formula, we obtain
$$(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3) = \lambda^3 - a\lambda^2 + b\lambda - 1$$
Notice the three factors in $c$ are simply $a - \lambda_i$ for $i = 1,2,3$. As a result,
$$c = (a-\lambda_1)(a-\lambda_2)(a-\lambda_3) = a^3 - a a^2 + b a - 1 = ba - 1$$
