Solve the cubic equation $z^3+6z^2+12z+16=0$
and show the three solutions on an Argand diagram
HINT: $(a+b)^3$
Solve the cubic equation $z^3+6z^2+12z+16=0$
and show the three solutions on an Argand diagram
HINT: $(a+b)^3$
Okay, lets proceed with your hint then.
$$z^3+6z^2+12z+16=0$$ $$(z+2)^3+8=0$$ $$(z+4)(z^2+2z+4)=0$$ $$(z+4)[(z+1)^2+3]=0$$
So you obtain $z=-4, -1+i\sqrt{3},-1-i\sqrt{3}$
It should be easy to represent these on an argand diagram.