To prove: x * y * z - xyz = 2

Let's start by taking the logarithm of both sides of each equation to simplify the expressions:

1. Taking logarithm of a^x = bc:
log(a^x) = log(bc)
x * log(a) = log(b) + log(c)

2. Taking logarithm of b^y = ac:
log(b^y) = log(ac)
y * log(b) = log(a) + log(c)

3. Taking logarithm of c^z = ab:
log(c^z) = log(ab)
z * log(c) = log(a) + log(b)



Now, let's solve these equations step by step:

1. Dividing equation (1) by equation (2):
(x * log(a)) / (y * log(b)) = (log(b) + log(c)) / (log(a) + log(c))
x / y = (log(b) + log(c)) / (log(a) + log(c))

2. Dividing equation (2) by equation (3):
(y * log(b)) / (z * log(c)) = (log(a) + log(c)) / (log(a) + log(b))
y / z = (log(a) + log(c)) / (log(a) + log(b))

3. Dividing equation (3) by equation (1):
(z * log(c)) / (x * log(a)) = (log(a) + log(b)) / (log(b) + log(c))
z / x = (log(a) + log(b)) / (log(b) + log(c))



Now, let's combine these equations to obtain a single equation:

(x / y) * (y / z) * (z / x) = [(log(b) + log(c)) / (log(a) + log(c))] * [(log(a) + log(c)) / (log(a) + log(b))] * [(log(a) + log(b)) / (log(b) + log(c))]
x * y * z / (x * y * z) = (log(b) + log(c))^2 / (log(a) + log(b))^2
1 = (log(b) + log(c))^2 / (log(a) + log(b))^2

Taking the square root of both sides:
1 = (log(b) + log(c)) / (log(a) + log(b))

Simplifying the equation:
(log(a) + log(b)) = log(b) + log(c)
log(a) = log(c)

Taking the exponential of both sides:
a = c



Now, substituting a = c into any of the original equations, let's use equation (1):

a^x = bc
(c)^x = bc
c^x = bc

Comparing this equation with equation (3):
c^z = ab

Since c^x = c^z, we can conclude that x = z