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

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1/(x^3 - 1)

We know that:

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

Then:

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

Let us rewrite:

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

= [1/(x-1)]*[1/(x^2 + x+ 1)]

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

To determine the partial fractions of 1/(x^3-1).

We knoe that x^3 -1 = (x-1)(x^2+x+1).

We suppose 1/(x^3-1) = a/(x-1) + (bx+c)/(x^2+x+c)....(1) And determine the constants a, b and c.

We multiply by x^3-1 both sides:

1 = a(x^2+x+1) + (bx+c)(x-1)......(2)

Put x = 1 in (1):

1 = a(1^2+1+1).

1 =3a.

So a = 1/3.

Put x = 0.

1 = (1/3)*1 -c. .

c = 1/3 -1 = -2/3.

Now equate x^2 terms on both sides:

ax^2+bx^2 = 0.

Therefore b = -a.

Or b = -a = -1/3.

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