2015/047) Prove that if $x = log_a(bc),y = log_b(ca),z = log_c(ab)$ then the value of $xyz - x - y - z$ is 2

we have

$x= log_a(bc)$

so $1 + x = 1 + log_a(bc) = log_a(a) + log_a(bc) = log_a(abc)$

or $\dfrac{1}{1+x} = log_{abc}(a) \cdots (1)$

similarly

$\dfrac{1}{1+y} = log_{abc}(b) \cdots (2)$

$\dfrac{1}{1+z} = log_{abc}(c) \cdots (3)$

 

 Adding (1), (2) and (3) we get
$\dfrac{1}{1+x}+ \dfrac{1}{1+y}+ \dfrac{1}{1+z} = log_{abc}(a) + log_{abc}(b) + log_{abc}(c) = log_{abc}(abc) = 1$

or
$(1+y)(1+z) + (1+z)(1+x) + (1+x)(1+y) = (1+x)(1+y)(1+ z)$

or

$1 + yz + y + z + 1 + xz + x + z + 1 + xy + x + z = 1 + x + y + z + xy + yz + zx + xyz$

or $2 + x + y+ z = xyz$

or $xyz – x – y -z = 2$