Find real roots of the equation

Find all real solutions to

$$\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} = x^3-x^2-x+1$$

This question is very similar to one of my previous problem, except I cannot find a monotonic function as has been done in the solution to that problem.

Any help will be appreciated.
Thanks.

This really is similar in some way to your previous problem.

$$\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} = x^3-x^2-x+1$$

Firstly, for all the terms to be valid, we do domain checking. Solving $x+1\ge0$, $2-x\ge0$, $x^2-x+2\ge0$ and $-x^2+x+1\ge0$, we arrive at $x\in[-0.618, 1.618]$ (actually from the last inequality only)

Putting $x+1:=\alpha$ and $x^2-x+2:=\beta$.

The left hand side is $$\frac{\sqrt{\alpha}}{2+\sqrt{3-\alpha}} - \frac{\sqrt{\beta}}{2+\sqrt{3-\beta}} $$

Now, consider the function $$f(x)=\frac{\sqrt{x}}{2+\sqrt{3-x}}$$ Indeed it is monotonic. (Check yourself) This implies that $f(\beta)\ge f(\alpha)$ iff $\beta\ge\alpha$. Now see that $\beta-\alpha=x^2-2x+1=(x-1)^2\ge0$ thus $\beta\ge\alpha$ and $f(\beta)\ge f(\alpha)$. Which means, $$f(\alpha)-f(\beta)=\dfrac{\sqrt{x+1}}{2+\sqrt{2-x}} - \dfrac{\sqrt{x^2-x+2}}{2+\sqrt{-x^2+x+1}} \le 0$$

So the Left hand side is less than or equal to zero.

As for the right hand side, it can be decomposed to $$x^3-x^2-x+1=x^2(x-1)-1(x-1)=(x-1)^2(x+1)$$ Since $(x-1)^2\ge0$, it's behaviour is controlled by $(x+1)$ and it is $0$ at $x=1,-1$ and negative in $x<-1$ which is outside our domain.

So we see that in our domain, $[-0.618, 1.618]$, the Left side is non-positive and the Right side is non-negative. The only possibility for a solution is if they both equal $0$. Hence, solving $\alpha=\beta\implies x+1=x^2-x+2$ and $x^3-x^2-x+1=0$, we get $x=1$.