Question 32328
[(7-x)/(x+3)][(x^3+27)/(x^2-9x+14)]
=[(7-x)/(x+3)][(x+3)(x^2-3x+9)/(x-7)(x-2)]
=[-(x-7)/(x+3)][(x+3)(x^2-3x+9)/(x-7)(x-2)]
= -(x^2-3x+9)/(x-2)
Answer: -(x^2-3x+9)/(x-2)