Solve the equation, indicate any extraneous solutions.

(2x)/(x^2-1)= (4x^2+6x-6)/(x^3+x^2-x-1) - (1)/(x+1)  simplify

[2x] / [ (x + 1) (x - 1) ]  =  [ 2 (2x^2 + 3x -  3)] / [ x^2 ( x + 1) - 1 (x + 1)] - (1) /(x + 1)

[2x] / [ (x + 1) (x - 1) ]  =  [ 2 (2x^2 + 3x -  3)] / [ (x^2 - 1) (x + 1)] - (1) / ( x + 1)

[2x] / [ (x + 1) (x - 1) ]  =  [ 2 (2x^2 + 3x -  3)] / [ (x - 1) (x + 1) (x + 1)] - (1) / ( x + 1)

Multiply through by the common denominator of (x - 1) (x + 1) (x + 1)

2x (x + 1)   = [2 (2x^2 + 3x - 3 ) ]  -  [(x - 1) (x + 1) ]    simplify

2x ^2 + 2x   = (4x ^2 + 6x - 6)  - [x^2 - 1) ]

2x^2 + 2x  = 3x^2 + 6x - 5

x^2 + 4x - 5   = 0     factor

(x + 5) (x - 1)   =  0      set both factors to 0

x + 5  = 0    →   x = -5   is one possible solution

x - 1  = 0  →    x = 1   is another possible solution

However........we must reject the second solution because it causes two of the denominators in the original problem to = 0 

So.......the only solution is    x  = -5

Is (x+1) (x+1) (x-1) the least common denominator?? Cant I just multiply everything by (x+1) ?

greeting Omi 67