<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] Putting another answer here, a simpler one. We start by multiplying both sides of the equation by $x\sqrt{x^2-1}$. This gives $$(x^4-3x^2+x+1)\sqrt{x^2-1}+(x^5-4x^3+x^2+3x-1)=0$$
Subtracting this from the original equation, we get directly:
$$\sqrt{x^2-1}=x^3-2x+1$$
We can see that $x=1$ is a solution and other solution must be for $|x|>1$.
We now look at $f(x)=x^3-2x+1-\sqrt{x^2-1}$. Its derivative is $f'(x)=3x^2-2-\frac x{\sqrt{x^2-1}}$.
For $x<-1$, $f'(x)$ always positive, so there is only one solution, which is $x=-\sqrt2$.
For $x>1$, we check that $f''(x)$ is always positive. So we can have only 2 solutions. We already have $x=1$ so the only one left is $x=\sqrt2$.
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