Đặt \(x^2-2=a;4x-6=b\)
Theođề, ta có \(\left(a-b\right)^3-a^3+b^3=0\)
\(\Leftrightarrow a^3-b^3+3ab\left(a-b\right)-a^3+b^3=0\)
=>3ab(a-b)=0
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(2-x^2\right)\cdot\left(4x-6\right)=0\)
hay \(x\in\left\{2;\sqrt{2};-\sqrt{2};\dfrac{3}{2}\right\}\)