Correct question is:

Find the value of x if

tan^-1 (x - 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1 (3x)

Solution: Given, tan^-1 (x - 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1 (3x)

=> tan^-1 (x - 1) + tan^-1 (x + 1) + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{(x - 1) + (x + 1)}/{1 - (x - 1) * (x + 1)}] + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{2x}/{(1 - (x² - 1)}] + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{2x}/{(1 - (x² - 1)}] + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{2x}/{(1 - x² + 1)}] + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{2x}/{(2 - x² )}] + tan^-1 x = tan^-1 (3x)

=> tan^-1 [{2x/(2 - x² ) + x}/{1 - 2x² /{(2 - x² )}]  = tan^-1 (3x)

=> tan^-1 [{2x + x(2 - x² )}/(2 - x² )]/[{1 - 2x² /{(2 - x² )}] = tan^-1 (3x)

=> tan^-1 [{2x + x(2 - x² )}/{(2 - x² ) - 2x² }] = tan^-1 (3x)

=> tan^-1 [{2x + 2x - x³ )}/{2 - x²  - 2x² }]  = tan^-1 (3x)

=> tan^-1 [{4x - x³ }/{2 - x²  - 2x² }] = tan^-1 (3x)

=> tan^-1 [{4x - x³ }/{2 - 3x² }] = tan^-1 (3x)

=> {4x - x³ }/{2 - 3x² } = 3x

=> 4x - x³ = {2 - 3x² ) * 3x

=> 4x - x³ = 6x - 9x³

=> 4x - x³ - 6x + 9x³ = 0

=> 8x³ - 2x = 0

=> 2x(4x² - 1) = 0

=> 2x*(2x - 1)*(2x + 1) = 0

=> x = 0, 1/2, -1/2

So, the value of x are 0, 1/2, -1/2