<s>[INST] 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 $x=\pm\sqrt{2}$.
 [/INST] Write $P\equiv \sqrt{x^2-1}$ and $Q\equiv x^3-3x+1$ then the first equation can be written as
$$
Q(x+P)+1=0 \tag{1}
$$
It's clear that $|x|\geq 1$ otherwise there is no solution for $P$. By inspection, $x=1$ is one of the solutions.
Assuming $|x|>1$, multiply $(1)$ by $(P-x)$:
$$
Q(P^2-x^2)+P-x=0
$$
Since $P^2-x^2=-1$,
$$
P=Q+x=x^3-2x+1
$$
Define $y=x^2-2$ then this becomes
$$
P=yx+1.
$$
Squaring,
$$
P^2\equiv y+1=(yx+1)^2
$$
and so
$$
y+1=(yx+1)^2
$$
If $x>1$, you can easily convince yourself that the only way for this to be true is if $y=0$, hence $x=\pm \sqrt2$.
However, there is a flaw that I have not yet convinced myself that if $x<-1$ there is no other obvious solution than $x=-\sqrt2$. Sorry to leave a slightly incomplete answer.
</s>