# absolute value of fractionthe problem gives the fraction (a^3+2a^2-a-2)/(a^3-2a^2-a+2) and the absolute value of fraction, 1. the problem requests a

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You also may form the following groups to ease the process of factorization, such that:

`(a^3 - a) + (2a^2 - 2) = a(a^2 - 1) + 2(a^2 - 1)`

Factoring out `a^2 - 1` yields:

`(a^3 - a) + (2a^2 - 2) = (a^2 - 1)(a + 2)`

`(a^3 - a) - (2a^2 - 2) = a(a^2 - 1) - 2(a^2 - 1)`

Factoring out `a^2 - 1` yields:

`(a^3 - a) - (2a^2 - 2) = (a^2 - 1)(a - 2)`

Hence, replacing the factorized form for the original forms in the given fraction, yields:

`((a^2 - 1)(a + 2))/((a^2 - 1)(a - 2))`

Reducing duplicate factors yields:

`((a^2 - 1)(a + 2))/((a^2 - 1)(a - 2)) = (a + 2)/(a - 2)`

The problem provides the information that the absolute value of the fraction equals 1, such that:

`|(a + 2)/(a - 2)| = 1`

Using the definition of absolute value yields:

`|(a + 2)/(a - 2)| = {((a + 2)/(a - 2), (a + 2)/(a - 2) >= 0),(-(a + 2)/(a - 2), (a + 2)/(a - 2) < 0):}`

`(a + 2)/(a - 2) = 1 => a + 2 = a - 2 => 2 = -2` invalid

`-(a + 2)/(a - 2) = 1 => -a - 2 = a - 2 => -a - a = 2 - 2 => -2a = 0 => {(-2!=0),(a=0):}`

Since the equation `-(a + 2)/(a - 2) = 1` is valid for` (a + 2)/(a - 2) < 0 => a in (-2,2)` , hence `a = 0` is valid.

**Hence, evaluating the solution to the given equation, under the given conditions, yields **`a = 0.`

We'll factorize the numerator in this way:

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

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

Now, we'll factorize the denominator:

(a^3-2a^2)-(a-2) = a^2(a-2) - (a-2)

a^2(a-2) - (a-2) = (a-2)(a^2-1)

We'll re-write the given fraction:

(a+2)(a^2-1)/(a-2)(a^2-1)

We'll simplify and we'll get:

(a+2)/(a-2)

We know, from enunciation, that;

|(a+2)/(a-2)| = 1

case 1)

(a+2)/(a-2) = 1

(a+2) = (a-2)

2 = -2 impossible

case 2)

(a+2)/(a-2) = -1

a+2 = -a + 2

2a = 0

**a = 0**

The value of a is 0 for the absolute value of the given fraction to be 1.