Answer
$$x = \pm \frac{1}{{\sqrt 3 }},{\text{ }}x = \pm \sqrt 2 $$
Work Step by Step
$$\eqalign{
& 3{x^6} - {x^4} - 12{x^2} + 4 = 0 \cr
& {\text{Grouping terms}} \cr
& \left( {3{x^6} - {x^4}} \right) - \left( {12{x^2} - 4} \right) = 0 \cr
& {\text{Factoring}} \cr
& {x^4}\left( {3{x^2} - 1} \right) - 4\left( {3{x^2} - 1} \right) = 0 \cr
& \left( {3{x^2} - 1} \right)\left( {{x^4} - 4} \right) = 0 \cr
& \left( {3{x^2} - 1} \right)\left( {{x^2} + 2} \right)\left( {{x^2} - 2} \right) = 0 \cr
& {\text{Zero - factor property}} \cr
& 3{x^2} - 1 = 0,{\text{ }}\underbrace {{x^2} + 2 = 0}_{{\text{No real solutions}}},{\text{ }}{x^2} - 2 = 0 \cr
& {x^2} = \frac{1}{3},{\text{ }}{x^2} = 2 \cr
& x = \pm \sqrt {\frac{1}{3}} ,{\text{ }}x = \pm \sqrt 2 \cr
& x = \pm \frac{1}{{\sqrt 3 }},{\text{ }}x = \pm \sqrt 2 \cr} $$
