Answer
$$x = \pm \frac{1}{{\sqrt 2 }},{\text{ }}x = \pm 1$$
Work Step by Step
$$\eqalign{
& 2{x^6} - {x^4} - 2{x^2} + 1 = 0 \cr
& {\text{Grouping terms}} \cr
& \left( {2{x^6} - {x^4}} \right) - \left( {2{x^2} - 1} \right) = 0 \cr
& {\text{Factoring}} \cr
& {x^4}\left( {2{x^2} - 1} \right) - \left( {2{x^2} - 1} \right) = 0 \cr
& \left( {2{x^2} - 1} \right)\left( {{x^4} - 1} \right) = 0 \cr
& \left( {2{x^2} - 1} \right)\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& 2{x^2} - 1 = 0,{\text{ }}\underbrace {{x^2} + 1 = 0}_{{\text{No real solutions}}},{\text{ }}{x^2} - 1 = 0 \cr
& {x^2} = \frac{1}{2},{\text{ }}{x^2} = 1 \cr
& x = \pm \sqrt {\frac{1}{2}} ,{\text{ }}x = \pm \sqrt 1 \cr
& x = \pm \frac{1}{{\sqrt 2 }},{\text{ }}x = \pm 1 \cr} $$
