Question

(x+1) (x+2) (x-3) = (x-2) (x+1) (x+1)

Answer 1

(x+1) (x+2) (x-3) = (x-2) (x+1) (x+1)

$\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x-2\right)\left(x+1\right)\left(x+1\right)$


$\left(x^2+3x+2\right)\left(x-3\right)=\left(x-2\right)\left(x+1\right)\left(x+1\right)$

$x^3-7x-6=\left(x-2\right)\left(x+1\right)\left(x+1\right)$

$x^3-7x-6=\left(x-2\right)\left(x+1\right)^2$

$x^3-7x-6=\left(x-2\right)\left(x^2+2x+1\right)$

$x^3-7x-6=x^3-3x-2$

$\mathrm{Sumar\:}6\mathrm{\:a\:ambos\:lados} x^3-7x-6+6=x^3-3x-2+6$

$x^3-7x=x^3-3x+4$

$\mathrm{Restar\:}x^3-3x\mathrm{\:de\:ambos\:lados} x^3-7x-\left(x^3-3x\right)=x^3-3x+4-\left(x^3-3x\right)$

$-4x=4$

$\frac{-4x}{-4}=\frac{4}{-4}$

$x=-1$

Answer 2

$(x+1) (x+2) (x-3) = (x-2) (x+1) (x+1)\\ \\ (x+1) (x+2) (x-3) - (x-2) (x+1) (x+1)=0\\ \\ (x+1)[(x+2) (x-3)-(x-2) (x+1)]=0\\ \\ (x+1)[(x^2-x-6)-(x^2-x-2)]=0\\ \\ (x+1)[-6+2]=0\\ \\ x+1=0\\ \\ \boxed{x=-1}$