Because of the factors `z-1` and `z+1,` we know that `+-1` are solutions.

Now we can use the quadratic formula for the remaining factors. For

`z^2+z+1,` we get

`z=(-1+-sqrt(1^2-4(1)(1)))/2=(-1+-i sqrt(3))/2`

and for `z^2-z+1` we get

`z=(1+-i sqrt(3))/2` .

**The six solutions are `+-1,(-1+-i sqrt(3))/2,(1+-i
sqrt(3))/2.`**

Note that if `z` is a solution, then so is ``the conjugate of `z.` This is true for any solution of `z^n=1` for any `n.` Also, if `z` is a solution, so is `-z.` This is the case for the solutions of `z^n=1` only when `n` is even (which is why it works for this example).

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**Further Reading**