Let each formula = 1, take the logarithm, and then add it to get:

(x+y+z)(lna +lnb+lnc)=0 so x+y+z=0 or lna+lnb+lnc = 0, that is, abc= 1.

If abc= 1 and A not = 1, then c= 1/(ab).

So you can get the original formula:

a^(x-z)*b^(y-z)= 1

a^(y-x)*b^(z-x)= 1

a^(z-y)*b^(x-y)= 1

It is easy to draw from the above three formulas that if two of x, y, z, y and z are equal, then the third one must be equal to the first two.

We assume that if x, y, z, y and z are not equal, there will be contradictions.

Let b = a t, then there are:

a^(x-z+t(y-z))= 1，= = & gtx-z=-t(y-z)

a^(y-x+t(z-x))= 1 == >y-x=-t(z-x)

a^(z-y+t(x-y))= 1 == >z-y=-t(x-y)

The three formulas on the right are multiplied: (x-z) (y-x) (z-y) =-t3 (y-z) (z-x) (x-y).

==》 t^3= 1，= = & gtt= 1

So the above three equations become:

x+y=2z

y+z=2x

z+x=2y

= = & gtX=y=z contradicts the assumption that x, y, z y and z are not equal to each other.

So the conclusion is established.