Let $$f(x) = (x+3)^{2}(x-2)^3$$, $$x \in [-4, 4]$$. If $$M$$ and $$m$$ are the maximum and minimum values of $$f$$, respectively in $$[-4, 4]$$, then the value of $$M - m$$ is:

 $$f(x) = (x+3)^2(x-2)^3$$ on $$[-4, 4]$$.

$$f'(x) = 2(x+3)(x-2)^3 + 3(x+3)^2(x-2)^2 = (x+3)(x-2)^2[2(x-2)+3(x+3)] = (x+3)(x-2)^2(5x+5) = 5(x+3)(x-2)^2(x+1)$$

Critical points: $$x = -3, -1, 2$$.

$$f(-4) = (-1)^2(-6)^3 = -216$$. $$f(-3) = 0$$.

$$f(-1) = (2)^2(-3)^3 = 4(-27) = -108$$. $$f(2) = 0$$. $$f(4) = (7)^2(2)^3 = 49(8) = 392$$.

Maximum $$M = 392$$ at $$x = 4$$. Minimum $$m = -216$$ at $$x = -4$$.

$$M - m = 392 - (-216) = 608$$.