<s>[INST] 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
$$ 3 = 0$$
 [/INST] 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 $\Bbb R$...
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