Extraneous solution when solving $x^2+x+1=0$ by getting $x^2=1/x$ Let's assume that we have$$x^2+x+1=0.\tag1$$
Substituting $x=0$, we get $1=0$, so $0$ is not a root for the quadratic equation and thus, $x\neq0$. Therefore, there exists $\frac{1}{x}$, which we'll multiply by both sides of $(1)$, giving us $$x+1+\frac{1}{x}=0.$$
We will, then, move $\frac{1}{x}$ to the other side and get $$x+1=-\frac{1}{x}.$$ If we add $x^2$ to both sides and note that $x^2+x+1=0$, we will have $x^2-\frac{1}{x}=0$. The real root of this equation is $x=1$, which is not a root of $(1)$.
I was wondering at which step did I do something that was incorrect and resulted in this supposed root.
 A: Hint:
$x^2+x+1=0$
is true only for some value of $x$ not $\forall x$.
A: To go from $x^2+x+1+\frac{1}{x}=x^2$ to $\frac{1}{x}=x^2$ you are also assuming $x^2+x+1=0$. So you are changing the single equation $x^2+x+1+\frac{1}{x}=x^2$ to the system of equations
$$\begin{align*}
x^2+x+1&=0\\ 
x^2&=\frac{1}{x}.
\end{align*}$$
This is equivalent to the system
$$\begin{align*}
x^2+x+1&=0\\
x^3&=1.
\end{align*}$$
This is, in turn, equivalent to the system
$$\begin{align*}
x^2+x+1&= 0\\
(x-1)(x^2+x+1)&=0.
\end{align*}$$
But this system is equivalent to $x^2+x+1=0$... that is, your original equation.
The error (or rather, the non-reversible step) lies in "forgetting" about the condition $x^2+x+1=0$ which you are assuming to get to $x^2=\frac{1}{x}$; by dropping it explicitly, you end up with the equation $x^3-1=0$, or $(x-1)(x^2+x+1)=0$; this has the solutions $x^2+x+1=0$ (the original equation) plus the solution $x-1=0$ (the extraneous solution which was introduced when you forgot to keep the global condition that $x^2+x+1=0$).
A: The first step when we got an extraneous root is the equation $x^2 - \frac{1}{x} = 0$. It is a corollary of equation(1). But it's only corollary: it's not equivalent  to it.
