Let’s simplify the function:
(x²(x²-2x-3))/(x²(x²-5x+6))=((x-3)(x+1))/((x-3)(x-2)), except when x=0;
(x+1)/(x-2) except when x-3=0, that is, when x=3. x-2=0, that is, x=2, means we will be dividing by zero.
So we can cancel out the factors x² and x-3, except when x=0 and 3, and we can evaluate the function, except when x=2. So the boxes must contain 0, 3 and 2.