# How to solve $z^6 + z^3 + 1 = 0$?

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?

• Hint: substitute $w=z^3$. Nov 25, 2017 at 13:26
• Use the quadratic formula on your second step, then take cube roots.
– Ned
Nov 25, 2017 at 13:26
• 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.
– user228113
Nov 25, 2017 at 13:30

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 $1=e^{2n\pi i}$.