Answer
$$x = \pm 1,{\text{ }}x = - 2,{\text{ }}x = \frac{{1 \pm \sqrt {13} }}{2}$$
Work Step by Step
$$\eqalign{
& {\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^3} - {\left( {{x^2} - 1} \right)^3}{\left( {x + 2} \right)^2} = 0 \cr
& {\text{Factoring, the common factor is }}{\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^2} \cr
& {\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^2}\left[ {\frac{{{{\left( {{x^2} - 1} \right)}^2}{{\left( {x + 2} \right)}^3}}}{{{{\left( {{x^2} - 1} \right)}^2}{{\left( {x + 2} \right)}^2}}} - \frac{{{{\left( {{x^2} - 1} \right)}^3}{{\left( {x + 2} \right)}^2}}}{{{{\left( {{x^2} - 1} \right)}^2}{{\left( {x + 2} \right)}^2}}}} \right] = 0 \cr
& {\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^2}\left[ {x + 2 - \left( {{x^2} - 1} \right)} \right] = 0 \cr
& {\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^2}\left( {x + 2 - {x^2} + 1} \right) = 0 \cr
& {\left( {{x^2} - 1} \right)^2}{\left( {x + 2} \right)^2}\left( {3 + x - {x^2}} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& {x^2} - 1 = 0,{\text{ }}x + 2 = 0,{\text{ }}{x^2} - x - 3 = 0 \cr
& x = \pm 1,{\text{ }}x = - 2,{\text{ }}{x^2} - x - 3 = 0 \cr
& {\text{Solving }}{x^2} - x - 3{\text{ by the quadratic formula}} \cr
& x = \frac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 1 \right)\left( { - 3} \right)} }}{{2\left( 1 \right)}} \cr
& x = \frac{{1 \pm \sqrt {13} }}{2} \cr
& {\text{The solutions are:}} \cr
& x = \pm 1,{\text{ }}x = - 2,{\text{ }}x = \frac{{1 \pm \sqrt {13} }}{2} \cr} $$
