What's wrong with this proof of $3=0$ starting from $x^2+x+1=0$? What am I missing?
$$ x^2 + x + 1 = 0$$
Then
$$ x^2 + x = -1$$
$$ x(x+1) = -1$$
$$ x(-x^2) = -1$$
$$x^3 = 1 $$
$$x = 1 $$
but
$$(1)^2 + (1) + 1 = 3 $$
So
$$ 3 = 0$$
 A: $x^3=1$ does not imply $x=1$. By doing "algebra" you can introduce false solutions which need to be checked.
$x^3=1 \iff (x-1)(x^2+x+1)=0$
You can see your original equation from here, with an additional solution added.
A: Nothing wrong: it is true that if $x^2+x+1=0$ and $x\in\Bbb R$, then $x=1$. Keeping all the pieces together $$x^2+x+1=0\Leftrightarrow\begin{cases}x^2+x=-1\\ x+1=-x^2\end{cases}\Leftrightarrow \begin{cases}-x^3=-1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ x+1=-x^2\end{cases}\Leftrightarrow\begin{cases}x=1\\ 2=-1\end{cases}$$
So the equation has no solutions in  $\Bbb R$.
Of course, we may argue that there is some degree of circularity in proving that $x^2+x+1$ has no real roots by using the fact that $x^3-1$ has exactly one root in $\Bbb R$...
A: If you are assuming $x$ is a real number, then all of these steps are valid, and prove that the assumption $x^2+x+1=0$ gives a contradiction.  In other words, this is just a proof that the equation $x^2+x+1=0$ has no real solutions.
If you allow $x$ to be a complex number, then $x^2+x+1=0$ does have solutions, but the step where you go from $x^3=1$ to $x=1$ is wrong: there are complex numbers besides $1$ whose cubes are $1$.  (This step is valid for real numbers, since the function $x\mapsto x^3$ from $\mathbb{R}$ to $\mathbb{R}$ is strictly increasing, so it can take the value $1$ at most once.)
A: You start with a solution to the equation $x^2 + x + 1 = 0$ and correctly show that it is also a solution to the equation $x^3 = 1$. You then observe that the only real solution to the second equation is $1$ (this is true).
You then conclude that $x = 1$ is a solution to the first equation (this is not true, there may be no real solutions to your original equation, as is in fact the case).
Analogy: if you started with $x^2 = -1$ and squared both sides, you would get $x^4 = 1$, but you can't conclude that $x = 1$.
