(27x^3 - 64)/(9x^2 - 16)

We notice that:

(27x^3 - 64) is difference between 2 cubes.

(9x^2 - 16) is a difference between 2 squares:

So, let us use the formula:

We kno wthat:

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

Then:

(27x^3 - 64) = (3x-4)(9x^2 + 12x + 16)

Also, we know that:

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

(9x^2 - 16) = (3x-4)(3x+4)

Now let us substitute:

(27x^3 - 64)/(9x^2 -16)=(3x-4)(9x^2 + 12x+16)/(3x-4)(3x+4)

Now eliminate similar terms:

**==> (27x^3-64)/(9x^2-16) = (9x^2 + 12x+16)/(3x+4)**

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

27x^3-64 = (3x)^3 - (4)^3

We'll apply the formula:

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

We'll put a = 3x and b = 4

(3x)^3 - (4)^3 = (3x-4)(9x^2 + 12x + 16)

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

9x^2-16 = (3x)^2 - 4^2

We'll apply the formula:

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

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

We'll substitute the differences by their products:

[(27x^3-64)/(9x^2-16)] = (3x-4)(9x^2 + 12x + 16)/(3x-4)(3x+4)

We'll simplify by the common factor (3x-4):

** [(27x^3-64)/(9x^2-16)] = [(9x^2 + 12x + 16)/(3x+4)]**

We can also combine the terms 12x + 16 and factorize them by 4;

[(27x^3-64)/(9x^2-16)] = 9x^2/(3x+4) + (12x + 16)/(3x+4)

[(27x^3-64)/(9x^2-16)] = 9x^2/(3x+4) + 4(3x + 4)/(3x+4)

We'll simplify the last ratio by (3x+4) and we'll get:

** [(27x^3-64)/(9x^2-16)] = 9x^2/(3x+4) + 4**