Question

solve for x... please give solution in detail​

Answer 1

Step-by-step explanation:

We have,

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

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

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

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

$\implies \frac{(x^{3} - 3x^{2} + 2x - 3 {x}^{2} + 9x - 6 ) - ( {x}^{3} + 3 {x}^{2} + 2x + 3 {x}^{2} + 9x + 6)}{(x + 1)(x + 2)(x + 3)} > 0$

$\implies \frac{(x^{3} - 6x^{2} + 11x - 6 ) - ( {x}^{3} + 6 {x}^{2} + 11x + 6)}{(x + 1)(x + 2)(x + 3)} > 0$

$\implies \frac{x^{3} - 6x^{2} + 11x - 6 - {x}^{3} - 6 {x}^{2} - 11x - 6}{(x + 1)(x + 2)(x + 3)} > 0$

$\implies \frac{ - 6x^{2} - 6 - 6 {x}^{2} - 6}{(x + 1)(x + 2)(x + 3)} > 0$

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

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

$\implies \frac{ (x^{2} + 1) }{(x + 1)(x + 2)(x + 3)} < 0$

Since, $x^2+1$ is always positive,

so,

$\implies (x + 1)(x + 2)(x + 3) < 0$

$\implies\:x\in(-\infty, -3)\:\cup\:(-2,-1)$