quotient of 64^3 - 27 and 4x-3

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

$$\\\frac{((4x)^3-3^3)}{4x-3}\\\\ =\frac{((4x)-3)((4x)^2+3*(4x)+3^2}{4x-3}\\\\ =\frac{(4x-3)(16x^2+12x+9)}{4x-3}\\\\ =16x^2+12x+9\qquad $Where x cannot = 3/4$$ $