First observe that the left hand side evaluates to 0 whenever [tex]x=-3[/tex], [tex]x=-2[/tex], or [tex]x=2[/tex].
So let's take numbers to either side of these roots. If [tex]x=-4[/tex], for instance, we have [tex](-4+3)^3=(-1)^3=-1<0[/tex], [tex](-4+2)^3=(-2)^3=-8<0[/tex], and [tex](-4-2)^2=(-6)^2=36>0[/tex]. We would be multiplying a positive number (2) by two negatives (-1 and -8) and a positive (36), which gives a positive number. So, we know [tex]x<-3[/tex] must be a solution set.
Doing the same for numbers between/beyond the other three cases, [tex]-3<x<-2[/tex], [tex]-2<x<2[/tex], and [tex]x>2[/tex], we would get, respectively, a negative, positive, and a positive.
The entire solution set would then be the union of the intervals over which we get a positive number, i.e. [tex](x<-3)\cup(-2<x<2)\cup(2<x)[/tex]
