Let's rewrite the inequality
#x^3-5x^2 < x-5#
#x^3-5x^2-x+5<0#
Let #f(x)=x^3-5x^2-x+5#
#f(1)=1-5-1+5 =0#
so,
#(x-1)# is a factor of #f(x)#
#f(-1)=-1-5+1+5=0#
so,
#(x+1)# is a factor of #f(x)#
So,
#(x+1)(x-1)=x^2-1# is a factor of #f(x)#
We perform a long division to find the 3rd factor
#color(white)(aaaa)##x^3-5x^2-x+5##color(white)(aaaa)##|##x^2-1#
#color(white)(aaaa)##x^3##color(white)(aaaaaa)##-x##color(white)(aaaaaaaa)##|##x-5#
#color(white)(aaaa)##0-5x^2##color(white)(aaa)##0+5#
#color(white)(aaaaaa)##-5x^2##color(white)(aaaaa)##+5#
#color(white)(aaaaaaaaa)##0##color(white)(aaaaa)##+0#
Therefore,
#f(x)=(x+1)(x-1)(x-5)#
Let's build the sign chart
#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##1##color(white)(aaaaa)##5##color(white)(aaaaa)##+oo#
#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#
#color(white)(aaaa)##x-5##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#
Therefore,
#f(x)<0# when #x in ]-oo, -1 [uu]1,5[#
