Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.

Since $$\tan\big(\arctan x+\arctan y\big)\equiv\frac{x+y}{1-xy}$$ and $\arctan$ has principal range $\left(-\frac\pi2,\frac\pi2\right),$ thus, for each $(x,y)$ and some $k\in\{-1,0,1\},$ $$\arctan x+\arctan y=\arctan\left(\frac{x+y}{1-xy}\right)+k\pi.$$
Therefore, \begin{align}&\arctan(x+1) + \arctan(x) + \arctan(x-1) = \arctan(3)\\
\implies&\arctan\left(\frac{x^3-4x}{3x^2-2}\right)=\arctan3+n\pi\quad\text{for some }n\in\{-2,-1,\ldots,2\}\\
\implies&\frac{x^3-4x}{3x^2-2}=3\\
\implies&\cdots\\
\implies&x=-1\:\:\text{or}\:\:5\pm\sqrt{19}\end{align}
$5-\sqrt{19}$ is the only non-extraneous solution. Desmos check.
