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 $x=3$. 

Consider the functions: $f_{1},f_{2}:D\rightarrow\textbf{R}$, where $D = (-\infty,-\sqrt{\frac{2}{3}}]\cup[\sqrt{\frac{2}{3}}, \infty) $,
$f_{1}(x)={\sqrt{3X^{2}-2}}+\sqrt[3]{x^{2}-1}$ and $f_{2}(x)=3x-2$.
For $x\leq-\sqrt{\frac{2}{3}}$ $f_{1}$ is decreasing, $f_{2}$ is increasing and $f_{1}(-\sqrt{\frac{2}{3}})>f_{2}(-\sqrt{\frac{2}{3}})$, the equation has no negative solutions.
For $x\geq\sqrt{\frac{2}{3}}$, $f_{1}$ is concave ( $f''_{1}(x) < 0$ ) and $f_{2}$ is linear function, the equation has at most two roots, which are $x_{1}=2$ and $x_{2}=3$. 
