Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
 & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.

Of course, $x= \pm \sqrt{2}$ are two roots of this equation. Assume $x^2 \ne 2$, we have:
\begin{align*}
&(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 &\Longleftrightarrow &  x^{3} - 3x + 1 = \sqrt{x^{2} - 1}-x \\\\
\Longleftrightarrow &  x^{3} - 2x + 1 = \sqrt{x^{2} - 1} & \Longleftrightarrow & x^{3} - 2x  = \sqrt{x^{2} - 1}-1\\\\
\Longleftrightarrow &  (x^{2} - 2)x  = \frac{x^2-2}{\sqrt{x^{2} - 1}+1} & \Longleftrightarrow & x = \frac{1}{\sqrt{x^{2} - 1}+1}
\end{align*}
Clearly, the only root for the last equation is $1$ because $LHS \ge 1 \ge RHS$.
So, the initial equation has exactly 3 roots $\{ 1, \sqrt{2},-\sqrt{2}\}$.
