# What is x for the fraction (4x^3-32)/[x^3+(x+2)^3] if not defined ?

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### 2 Answers

A fraction (4x^3-32)/[x^3+(x+2)^3] is not defined whenever the denominator is equal to 0.

x^3+(x+2)^3 = 0

=> x^3 + x^3 + 12x + 6x^2 + 8 = 0

=> 2x^3 + 6x^2 + 12x + 8 = 0

=> x^3 + 3x^2 + 6x + 4 = 0

=> x^3 + x^2 + 2x^2 + 6x + 4 = 0

=> x^3 + x^2 + 2x^2 + 4x + 2x + 4 = 0

=> x^2(x +1) +2x(x + 1) + 4(x +1) =0

=> (x^2 + 2x + 4)(x + 1) = 0

x1 = -1

x2 = [-2 + sqrt (4 - 16)] / 2

=> x2 = -1 + i*sqrt 12 / 2

=> x2 = -1 + i*sqrt 3

x3 = -1 - i*sqrt 3

We see that for the values x2 and x3, the fraction takes on the form 0/0.

**The values of x for which the fraction is not defined are -1 , -1 + i*sqrt 3 and -1 - i*sqrt 3**

A fraction is undefined if and only if it's denominator is cancelling.

So, we'll have to find out the roots of the denominator.

We notice that we can write the numerator as a difference of cubes, after factorizing by 4:

4(x^3 - 8) = 4(x - 2)(x^2 + 2x + 4)

We'll write the denominator as a sum of cubes:

x^3+(x+2)^3 = (x + x + 2)[x^2 - x(x+2) + (x+2)^2]

We'll expand the square and we'll remove the brackets inside the second factor's brackets:

x^2 - x(x+2) + (x+2)^2 = x^2 - x^2 - 2x + x^2 + 4x + 4

We'll combine and eliminate like terms:

x^2 - x(x+2) + (x+2)^2 = (x^2 + 2x + 4)

We'll re-write the fraction:

4(x - 2)(x^2 + 2x + 4)/(2x+2)(x^2 + 2x + 4)

We'll simplify:

4(x - 2)/2(x+1) = 2(x-2)/(x+1)

The fraction is undefined for x + 1 = 0

x = -1

**The value of x for the fraction if undefined is x = -1.**