[Crimson Boy Can You Solve This?] [0] easy algebra

This is an easy algebra problem. I believe clever jgh must can solve it in 1min!
We know that x-\frac{y}{z}-\frac{z}{y}=y-\frac{x}{z}-\frac{z}{x}=z-\frac{x}{y}-\frac{y}{x}=a and x {\neq}y{\neq}z
(1) What is the value of x+y+z ?
(2) What is the value of a ?
My hape solution is below the picture

This is my hape solution:
x-\frac{y}{z}-\frac{z}{y}-(y-\frac{x}{z}-\frac{z}{x})=0
(x-y)(1+\frac{1}{z}-\frac{z}{xy})=0
\frac{1+z}{z}=\frac{z}{xy}
Similarly \frac{1+x}{x}=\frac{x}{zy}
\frac{1+z}{1+x}=\frac{z^3}{x^3}
(1+z)x^3-z^3 x-z^3=0
Then we devide the upper thing by x-z because if x=z then the second upper thing must equal.
Finally we got (1+z)x^2+z(1+z)x+z^2=0
Similarly (1+z)y^2+z(1+z)y+z^2=0
So x+y=-\frac{(1+z)z}{1+z}=-z
x+y+z=0
xy=\frac{z^2}{1+z}
Then plug those two upper things into z-\frac{x}{y}-\frac{y}{x}=a and got a=1
I think mk must have a better solution!! If you have, please send another blog to let me see clearly! Thank you!