<s>[INST] Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:

*

*Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.


*Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$


*From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$


*Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.


*If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$
I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:

 He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?

 [/INST] You made a mistake When you said that if $$x^3=1$$
then $$x=1$$
In fact,
$$x^3=1\iff x^3-1=0\iff (x-1)(x^2+x+1)=0$$
$$\implies x=1 \text{ xor } x^2+x+1=0$$
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