Solve tan⁻¹(x - 1) + tan⁻¹x + tan⁻¹(x + 1) = tan⁻¹3x…

Step 1. The tangent of the left side. Let $a = x - 1$, $b = x$ and $c = x + 1$. Then $a + b + c = 3x$, $abc = x^3 - x$ and $ab + bc + ca = 3x^2 - 1$. So the tangent is $\dfrac{3x - x^3 + x}{1 - 3x^2 + 1} = \dfrac{4x - x^3}{2 - 3x^2}$.

Step 2. Set it equal to $3x$. $4x - x^3 = 6x - 9x^3$, so $8x^3 - 2x = 0$, that is $2x(4x^2 - 1) = 0$. The roots are $x = 0$ and $x = \pm\dfrac{1}{2}$.

Step 3. Check. At $x = 0$ both sides are $0$. At $x = \dfrac{1}{2}$ the left side is $-0.4636 + 0.4636 + 0.9828 = 0.9828 = \tan^{-1}1.5$, the right side. At $x = -\dfrac{1}{2}$ everything changes sign. All three are real solutions.