Note: 1/(x^3 -x)=1/x(x^2 - 1) = 1/x(x+1)(x-1) = A/(x-1) + B/x + C/(x+1)

The Lcm of A/(x-1) + B/x + C/(x+1) is x(x+1)(x-1), so we can rewrite A/(x-1) + B/x + C/(x+1) as [A(x^2 + x) + B(x^2 -1) + C(x^2 -x)]/x(x+1)(x-1)

Expanding the numerator we get:

(Ax^2 + Ax + Bx^2 - B + Cx^2 - Cx)/x(x+1)(x-1)

Grouping like powers of x we get:

[(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)

So now we have 1/x(x+1)(x-1) = [(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)

Since the denominators are equal, the numerators must be equal to maintain the equation.

=> 1 = (A+B+C)x^2 + (A-C)x - B

Now we equate like powers of x (Note we can think of 1 as 0x^2 + 0x +1)

Then A+B+C=0-------------(1)

A-C=0----------------(2)

B = -1 ----------------(3)

=> A+C=1 -------------------Plugged in the value of B

A-C=0 --------------------rewrite equation 2

Adding these equations eliminates C, so we end up with:

2A=1 => A=1/2.

But A+C = 1, => C=1/2.

Thus the constants are A=1/2, B=-1, C=1/2

So 1/(x^3 - x) = (1/2)/(x-1) - 1/(x) + (1/2)/(x+1), and you can check this.

There is another major method used to come up with the 3 equations, but this is the mainstream one I think.