What is the partial fraction descomposition of 1 / (x^6 - x^3) ?

Question

I think the partial fraction descomposition of 1 / (x^6 - x^3) is [ A/x + B/x + C/x + D/(x-1) + E/(x+1) + F/(x-1) ] is that correct?? yes/no?

Answer 1

No.

x^6-x^3=x^3 (x-1)(x^2+x+1)

It must be

1 / (x^6 - x^3) is A/x^3 + B/x^2 + C/x + D/(x-1) +( Ex+F)/(x^2+x+1)

Answer 2

No, that is not correct. The decomp is A/x + B/x^2 + C/x^3 +
D/(x-1) + (Ex+F)/(x^2+x+1).

Standard techniques show that A = B = 0, C = -1, D = 1/3,
E = -1/3, and F = -2/3 .

Answer 3

-(x + 2)/(3(x² + x + 1)) + 1/(3(x - 1)) - 1/x³