If x = b c, y = b a and z = _b a. Prove that xyz = xy + yz + zx

To prove the identity $x+y+z = xyz-2$, we will first express $x, y, z$ in a more expanded form using logarithm properties.

Step 1: Rewrite $x, y, z$ using the logarithm property $\log_p (qr) = \log_p q + \log_p r$.

$$x = \log_a b + \log_a c$$ $$y = \log_b c + \log_b a$$ $$z = \log_c a + \log_c b$$

Step 2: Introduce new variables for simplification.
Let $A = \log_a b$, $B = \log_b c$, and $C = \log_c a$.
Using the property $\log_p q = \frac{1}{\log_q p}$, we can write the other terms:

$$\log_a c = \frac{1}{\log_c a} = \frac{1}{C}$$ $$\log_b a = \frac{1}{\log_a b} = \frac{1}{A}$$ $$\log_c b = \frac{1}{\log_b c} = \frac{1}{B}$$

Step 3: Express $x, y, z$ in terms of $A, B, C$.

$$x = A + \frac{1}{C}$$ $$y = B + \frac{1}{A}$$ $$z = C + \frac{1}{B}$$

Step 4: Calculate the left-hand side (LHS) of the equation, $x+y+z$.

$$x+y+z = \left(A + \frac{1}{C}\right) + \left(B + \frac{1}{A}\right) + \left(C + \frac{1}{B}\right)$$ $$$$