`(2x)/(x^3 - 1)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

The denominator factors out as `x^3-1=(x-1)(x^2+x+1),` thus the general decomposition is

`(2x)/(x^3-1)=A/(x-1)+(Bx+C)/(x^2+x+1).`

To find A, B and C, multiply both sides by `x^3-1:`

`2x = A(x^2+x+1)+(Bx+C)(x-1),` or

`2x=x^2(A+B)+x(A+C-B)+(A-C),`

so

A+B=0, A+C-B=2 and A-C=0.

From this B=-A, C=A, A+C-B=A+A+A=3A=2, so A=2/3, B=-2/3, C=2/3.

Now check this:

`(2/3)*(1/(x-1)+(-x+1)/(x^2+x+1))=(2/3)*(x^2+x+1-x^2+2x-1)/(x^3-1)=`

`=(2/3)*(3x)/(x^3-1)=(2x)/(x^3-1),` which is correct.