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 $tan^{-1}(x) \pm \tan^{-1}(y)$.

I advise you to apply the identity
\begin{align*}
\tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - \tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)}
\end{align*}
so that you arrive at the following equation:
\begin{align*}
\frac{3x  - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3 & \Longleftrightarrow \frac{x^{3} - 4x}{3x^{2} - 2} = 3\\\\
& \Longleftrightarrow x^{3} - 9x^{2} - 4x + 6  = 0\\\\
& \Longleftrightarrow (x^{3} + x^{2}) - 10(x^{2} + x) + 6(x + 1) = 0\\\\
& \Longleftrightarrow x^{2}(x^{2} + 1) - 10x(x + 1) + 6(x + 1) = 0\\\\
& \Longleftrightarrow (x^{2} - 10x + 6)(x + 1) = 0
\end{align*}
Hence the solution set is a subset of $S = \{5\pm\sqrt{19},-1\}$.
EDIT
It remains to determine if there are extraneous roots.
In order to do so, you can check out WolframAlpha (for example).
