Logarithms will help you here ..

Substitute $c$ in $c^{z} = a$ for $b$ , then substitute $b$ for $a^{x}$ . We get $a^{xyz} = a$ , and from this we can get $xyz = log_{a}a$ , which is just $\boxed{1}$ .

Take ln of both sides of each equation, and bring exponents to front.

x ln a = ln b , y ln b = ln c , z ln c = ln a.

So, xyz = ln b/ ln a * ln c / ln b * ln a / ln c = 1 by cancellation.

This problem is wrong. Take a=0,b=0 and c=0.

Then x,y and z can be any number besides 0.

Also, take a=-1,b=-1 and c=-1.

Then x,y and z can be any odd integers.

such a simple problem...

c^z=a ---->(c^z)^x=a^x- since a^x=b

so c^(x z)=b--------<[c^(z x)]^y=b^y=c

thus c^(x y z)=c

therefore x y z=1