Before continuing, let's note that as #x, y, z > 0# we have #a, b, c >0# and so we may take square roots of any product or quotient thereof, and do so without considering negative solutions.

#{(xy = a),(yz = b),(xz=c):}#

From the first equation, we have

#y = a/x#

Substituting this into the second equation:

#b = yz = a/xz#

#=> z = b/ax#

Substituting this into the third equation:

#c = xz = b/ax^2#

#=> x^2 = (ac)/b#

#=> x = sqrt((ac)/b)#

Substituting our solution for #x# into our intermediate result from working on the first equation:

#y =a/x = a/sqrt((ac)/b) = sqrt((ab)/c)#

Substituting our solution for #x# into our intermediate result from working on the second equation:

#z = b/ax = b/asqrt((ac)/b) = sqrt((bc)/a)#

#:.{(x = sqrt((ac)/b)), (y = sqrt((ab)/c)), (z = sqrt((bc)/a)):}#