Given: $f\left(x\right)={\left(x+3\right)}^2{\left(x-2\right)}^3$

$\Rightarrow f^{\text{'}}\left(x\right)={\left(x+3\right)}^2\times \left(3\right){\left(x-2\right)}^2+{\left(x-2\right)}^3\left(2\right)\left(x+3\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=\left(x+3\right){\left(x-2\right)}^2\left[\left(x+3\right)\times \left(3\right)+\left(x-2\right)\left(2\right)\right]$

$\Rightarrow f^{\text{'}}\left(x\right)=\left(x+3\right){\left(x-2\right)}^2\left[3x+9+2x-4\right]$

$\Rightarrow f^{\text{'}}\left(x\right)=\left(x+3\right){\left(x-2\right)}^2\left(5x+5\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=5\left(x+3\right){\left(x-2\right)}^2\left(x+1\right)$

For critical points, we put $f^{\text{'}}\left(x\right)=0$.

$\Rightarrow 5\left(x+3\right){\left(x-2\right)}^2\left(x+1\right)=0$

$\Rightarrow x=-3, 2, -1$

$\Rightarrow f\left(-3\right)=f\left(2\right)=0$

$\Rightarrow f\left(-1\right)=2^2{\left(-3\right)}^3=-108$

Also, $x\in \left[-4,4\right]$

$\Rightarrow f(-4)=-216$

$\Rightarrow f\left(4\right)=49\times 8=392$

$\Rightarrow M=392, m=-216$

$\Rightarrow M-m=392+216$

$\Rightarrow M-m=608$