# Finding Cubic Roots of a polynomial and using Argand Diagrams

Solve the cubic equation $z^3+6z^2+12z+16=0$

and show the three solutions on an Argand diagram

HINT: $(a+b)^3$

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Is that a question? What are your thoughts? – Henning Makholm Dec 30 '11 at 13:27
@David, the OP's own hint, namely recognizing the binomial expansion of $(z+2)^3+8=0$, seems to be easier than dividing out one of the solutions. – Henning Makholm Dec 30 '11 at 13:39
@Henning Makholm Indeed... Deleting my previous comments. – David Mitra Dec 30 '11 at 13:45

$$z^3+6z^2+12z+16=0$$ $$(z+2)^3+8=0$$ $$(z+4)(z^2+2z+4)=0$$ $$(z+4)[(z+1)^2+3]=0$$
So you obtain $z=-4, -1+i\sqrt{3},-1-i\sqrt{3}$
But you might miss an important feature of the diagram if you do it this way. I'd go $x+2=-2Q$, where $Q$ is any of the three cube roots of 1. Now if you know the geometry of the cube roots of 1, you can plot them (without having to know anything about how big $\sqrt3$ is), then multiply by $-2$, then shift left by 2. – Gerry Myerson Dec 30 '11 at 15:03