Proof: Consider the quadratic equation $x^2+x+1=0$. Then, we can see that $x^2=−x−1$. Assuming that $x$ is not zero (which it clearly isn't, from the equation) we can divide by $x$ to give
$$x=−1−\frac{1}{x}$$
Substitute this back into the $x$ term in the middle of the original equation, so
$$x^2+(−1−\frac{1}{x})+1=0$$
This reduces to
$$x^2=\frac{1}{x}$$
So, $x^3=1$, so $x=1$ is the solution. Substituting back into the equation for $x$ gives
$1^2+1+1=0$Therefore, $3=0$.
What happened?
Answer
Up to $x^3=1$ everything is fine. This allows you to conclude that $x\in \{z\in \Bbb C\colon z^3=1\}$. Since $\{z\in \Bbb C\colon z^3=1\}=\left\{\dfrac{-1 + i\sqrt 3}{2}, \dfrac{-1 - i\sqrt 3}{2},1\right\}$, then $x$ is one of the elements of this set.
You made a reasoning consisting of logical consequences, not one of logical equivalences. That's why you can tell that $x\in \{z\in \Bbb C\colon z^3=1\}$, but you can't say which one is it.
See this for a similar issue.
An even more simpler version of your mistake is this: suppose $x^2=1$, then $x=1$.
You can convince yourself that this is wrong and that you did the same thing in your question.
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