Question

Solve the following inequality:
$\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1$

Answer 1

$=>\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}>1$

$=>\frac{(x-1)(x-2)(x-3)-(x+1)(x+2)(x+3)}{(x+1)(x+2)(x+3)}>1$

$=>\frac{x^3-{3x}^2+2x-{3x}^2+9x-6-x^3-{3x}^2-2x-{3x}^2-9x-6}{(x+1)(x+2)(x+3)}>1$

$=>\frac{-12x^2-12}{(x+1)(x+2)(x+3)}>0$

$=>\frac{(x^2-1)}{(x^2+1)(x+2)(x-3)}<0$

$=>\frac{x-1}{(x+2)(x+3)}<0$

$=>xϵ(-\infty ,-3)\cup (-2,1)$.