Answer
$$x = 1,{\text{ }}x = \pm \sqrt 5 $$
Work Step by Step
$$\eqalign{
& {x^3} - {x^2} - 5x + 5 = 0 \cr
& {\text{Grouping terms}} \cr
& \left( {{x^3} - {x^2}} \right) - \left( {5x - 5} \right) = 0 \cr
& {\text{Factoring}} \cr
& {x^2}\left( {x - 1} \right) - 5\left( {x - 1} \right) = 0 \cr
& \left( {x - 1} \right)\left( {{x^2} - 5} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& x - 1 = 0,{\text{ }}{x^2} - 5 = 0 \cr
& x = 1,{\text{ }}x = \pm \sqrt 5 \cr} $$
