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?
Consider the equation $x=x^2$. Its solutions are easy to find: they are $0$ and $1$, right?!
However, if $x$ is a solution of that equation, then $x=x^2$, and therefore $x=(x^2)^2=x^4$. But the equation $x=x^4$ has other solutions, besides $0$ and $1$, namely $-\frac12+\frac{\sqrt3}2i$.
The problem lies in the if-then clause of the first sentence of the previous paragraph: we are saying (correctly) that if $x$ is a solution of the given equation, then $x$ is also a solution of another equation. But there is no reason for you to think that the new equation only has the solutions of the original one!
That's what happens in your example: if $x$ is a solution of $x^2+x+1=0$, then $x$ is a solution of $x^2=\frac1x$, but you have no reason to suppose that every solution of $x^2=\frac1x$ is also a solution of $x^2+x+1=0$.
Call $f(x)=x^2 + x + 1$, then his equation is $f(x)=0$.
From there, he derives the equation $(2)$, $\frac{f(x)}{x}=0$.
To reach equation $(3)$, he substracts the second equation to the first one, obtaining the equation $\left(\frac{1}{x}-1\right)f(x)=0$.
Thus, the equation $(3)$ will be true when either $f(x)=0$, which are the solutions of the original problem, or when $\frac{1}{x}-1=0$, that is, $x=1$. That´s what he means when he says he is adding the solution $x=1$.