# FunctionsFactor the rational function 1/(x^3-1) .

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You should use the following formula to factor denominator such that:

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

Hence, factoring out the fraction yields:

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

We have to factor 1/(x^3-1)

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

We cannot proceed further unless partial fractions are required.

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