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
$$ 3 = 0$$

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 $x = 1$.
