Question

Prove that if x = $lo{g}_{c}b+lo{g}_{b}c$
$y=lo{g}_{a}c+lo{g}_{c}a,$
$z=lo{g}_{b}a+lo{g}_{a}b$
then $xyz=x^2+y^2+z^2-4.$

Answer 1

Let A = $lo{g}_c$ c, b = $lo{g}_a$ c, C = $lo{g}_b$ a
$\therefore$ ABC = 1 clearly
x = $lo{g}_c$ b + $lo{g}_b$ c = A + $\frac{1}{A}$
$\therefore xyz=(A+\frac{1}{A})(B+\frac{1}{B})(C+\frac{1}{C})$
= $(A+\frac{1}{A})[BC+\frac{1}{BC}+\frac{B}{C}+\frac{C}{B}]$
= $(A+\frac{1}{A})[A+\frac{1}{A}+\frac{B}{C}+\frac{C}{B}]$ by (1)
= ${(A+\frac{1}{A})}^2+\left[(A+\frac{1}{A})(\frac{B}{C}+\frac{C}{B})\right]$
$\because A=\frac{1}{BC}$
= ${(A+\frac{1}{A})}^2[(B^2+\frac{1}{B^2})+(C^2+\frac{1}{C^2})]$
Add 2 + 2 in 2nd bracket and subtract 4 and put $A+\frac{1}{A}=x$ by (2)
$\therefore xyz=x^2+y^2+z^2-4.$