Let #x# be the cube root of #-1#, then we have #x*3=-1#

or #x^3+1=0#

#hArrx^3+x^2-x^2-x+x+1=0#

or #x^2(x+1)-x(x+1)+1(x+1)=0#

or #(x+1)(x^2-x+1)=0#

Hence either #x+1=0# i.e. #x=-1#, or #x^2-x+1=0#.

So one root is #x=-1# and for other two roots of #x^2-x+1=0#, we proceed as follows:

#x^2-x+1=0hArrx^2-2xx x xx (1/2)+(1/2)^2-(1/2)^2+1=0#

or #(x-1/2)^2+3/4=0#

i.e. #(x-1/2)^2-(sqrt3/2i)^2=0#

i.e. #(x-1/2+sqrt3/2i)(x-1/2-sqrt3/2i)=0#

i.e. #x=1/2-sqrt3/2i# or #x=1/2+sqrt3/2i#

Hence, three cube roots of #1# are #{-1,1/2-sqrt3/2i,1/2+sqrt3/2i}#