Can someone explain to me how to solve the equation $ z^6 + z^3 + 1 = 0 $ ?

I started working it out and obtained the following:
$ z^6 + z^3 = -1$
$ z^2 + z = -1 $
$ (x+iy)^2 + (x+iy) = -1 $
$ x^2 + 2xiy -y^2 = -1 $

And now I'm stuck. What would be my next step?

  • 3
    $\begingroup$ Hint: substitute $w=z^3$. $\endgroup$
    – user1337
    Nov 25, 2017 at 13:26
  • 1
    $\begingroup$ Use the quadratic formula on your second step, then take cube roots. $\endgroup$
    – Ned
    Nov 25, 2017 at 13:26
  • $\begingroup$ By the way, when you expanded the square, you forgot to add the linear term. Not that it matters, but be careful with your algebra. $\endgroup$
    – user228113
    Nov 25, 2017 at 13:30

4 Answers 4


If you multiply the equation by $z^3-1$ it becomes $z^9-1=0$. The answers you require are therefore the six ninth roots of unity that are not third roots of unity!


Start by writing $y = z^3$.

Then you get $y^2 + y + 1 = 0$.

Note that the two roots of this quadratic are precisely the two non-real cube roots of $1$. They can be represented as $\displaystyle \omega = e^{\frac{2\pi i}{3}}$ and $\displaystyle \omega^2 = e^{\frac{4\pi i}{3}}$. However, it is prudent to find a more general representation so that we don't miss out on any roots when we next take the cube root of these to solve for $z$.

So represent $\displaystyle y = e^{\frac{2k\pi i}{3}}, k \not\equiv 0\pmod 3$ (the restriction on $k$ is necessary to exclude the real cube root of unity, i.e. $1$ itself).

Then $z = e^{\frac{2k\pi i}{9}}, k \not\equiv 0\pmod 3$.

The unique solutions for $z$ are therefore: $\displaystyle z = e^{\frac{2\pi i}{9}}, e^{\frac{4\pi i}{9}}, e^{\frac{8\pi i}{9}}, e^{\frac{10\pi i}{9}}, e^{\frac{14\pi i}{9}}, e^{\frac{16\pi i}{9}}$


Let : $w=z^3$. Then your given equation becomes :

$$w^2 + w + 1 =0$$

which, using the quadratic formula for complex solutions, yields :

$$w = -\frac{1}{2} \pm \frac{\sqrt3}{2}i$$

Substituting back for $z^3$, can you calculate $z$ ? Be careful on the signs and on the amount of solution you must get !


There are two ways of getting into the original equation which can help you to solve this.

The first is to set $y=z^3$ and first solve $y^2+y+1=0$ and then find the cube roots of those solutions.

A second way is to note that if $z^3=1$, $z$ is not a solution of the equation. If you multiply the original by $z^3-1$ you will obtain $$z^9-1=0$$

You are therefore looking for the ninth roots of unity which are not cube roots of unity. Use the form $1=e^{2n\pi i}$.


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