Question 977750
((x+1)^3 (3x-5) - (x+1)^2 (8x+3))/(x-4)(x+1)^3=

((x+1)^3)(3x-5))/(x-4) (x+1)^3     minus     (x+1)^2(8x+3)/(x-4) (x+1)^3.   Split the fraction.


Cancel the (x+1)^3 completely in the first numerator term and (x+1)^2 in the second term.

This leaves 
[(3x-5)/(x-4)]  - ((8x+3)/(x+1)


common denominator is (x-4)(x+1)
numerator will be (3x-5)(x+1)=3x^2-17x20  -  [8x+3)(x-4)]
second part is 8x^2-29x-12,
3x^2-17x+20-8x^2+29x+12= {-5x^2+12x+32}/(x-4)(x+1)=  -(5x^2-12x-32)/(x-4)(x+1)=
-(5x+8)(x-4)/(x-4)(x+1)= -(5x+8)/(x+1), because the (x-4) divide out.

Notice that I pull a minus sign out of the equation, because factoring -x^2 terms is difficult.

Do not multiply out the numerator at first.  Split it into two fractions, cancel what you can, then put everything over a common denominator and simplify that.  Often you can cancel further.