(8x^3-27)/(4x^2-9)

To evaluate the expression we'll use factorization. We
notice that the numerator is a difference of cubes:

8x^3-27
= (2x)^3 - (3)^3

We'll apply the
formula:

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

We'll put a = 2x and b =
3

(2x)^3 - (3)^3 = (2x-3)(4x^2 + 6x +
9)

We also notice that the denominator is a difference of
squares:

4x^2-9 = (2x)^2 -
3^2

We'll apply the
formula:

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

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

We'll substitute the differences by their
products:

 [(8x^3-27)/(4x^2-9)] = (2x-3)(4x^2 + 6x +
9)/(2x-3)(2x+3)]

We'll simplify by the common factor
(2x-3):

 [(8x^3-27)/(4x^2-9)] =  [(4x^2 + 6x
+ 9)/(2x+3)]

We can also combine the terms
6x + 9 and factorize them by 3;

[(8x^3-27)/(4x^2-9)] =
4x^2/(2x+3) + (6x + 9)/(2x+3)

[(8x^3-27)/(4x^2-9)] =
4x^2/(2x+3) + 3(2x + 3)/(2x+3)

We'll simplify the last
ratio by (2x+3) and we'll
get:

[(8x^3-27)/(4x^2-9)] = [4x^2/(2x+3)] +
3