Solving $\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2 $ How can I solve the equation  $$\sqrt{3x^2-2}+\sqrt[3]{x^2-1}= 3x-2$$
I know that it has two roots: $x=1$ and $x=3$. 

Substituting $x = \sqrt{t^3+1}$ and twice squaring, we arrive to the equation
$$   36t^6-24t^5-95t^4+8t^3+4t^2-48t=0.$$ Its real roots are $t=0$ and $t=2$ (the latter root is found in the form $\pm \text{divisor}(48)/\text{divisor}(36)$), therefore $x=1$ and $x=3$.
