Factor the rational function 1/(x^3-1)

We notice that the denominator is a difference of cubes
and we'll re-write it using the formula:

a^3 - b^3 =
(a-b)(a^2 + ab + b^2)

We'll put a^3 = x^3 and b^3 =
1

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

We notice that the factor x^2 + x + 1 cannot be further
factored, since the discriminant of the quadratic is
negative.

delta = b^2 - 4ac, where a=1, b=1,
c=1

delta = 1 - 4

delta = -3
< 0

The final
factorized form of the given function
is:

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

We can
further decompose the factorized form of the rational function in the elementary
quotients.

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