Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found: 
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$ 
Therefore, it is legal to divide each term by $x$: 
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks. 
 A: You can indeed substitute. First, though, note that $1$ is not a solution to $x  = -1 - 1/x$.  So, by making that substitution, we are excluding $x = 1$ as a solution to our equation.  In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$. 
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$. 
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$. 
A: The higher level description of your work is:


*

*Assume $x$ is a solution to the original equation.

*Then $x$ has to be $1$

*$1$ is not a solution to the original equation.


And therefore we conclude the assumption is false: that is,


*

*Therefore, the original equation has no solutions.



Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to


*

*Assume $x$ is a solution to the original equation.

*Then $x$ has to be $1$ or either $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ (because those are the three cube roots of $1$)

*$1$ is not a solution to the original equation.

*$-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$ are solutions to the original equation


and therefore


*

*The solutions to the equation are $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$
A: Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$\begin{align}
x_1 &= 1, & \text{or} \\
x_2 &= -\frac{1}{2} + \frac{\sqrt 3}{2} i, & \text{or} \\
x_3 &= -\frac{1}{2} - \frac{\sqrt 3}{2} i
\end{align}$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this. 
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
A: For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $\,x=1\,$. We can substitute $\,1 \mapsto x\,$ on the RHS and end up with $\,x=x\,$. Of course that $\,x=1 \implies x=x\,$, but the converse is not true.
In OP's case, the original equation is quadratic in $\,x\,$ which has $2$ roots in $\Bbb C\,$, while the derived equation is a cubic which has $3$ roots in $\Bbb C\,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $\,x=1\,$ as noted already, which does not satisfy the original quadratic.
A: Transformations you apply to an equation may introduce alien solutions.
Taking an extreme example,
$$x=0\implies 0=0$$ which is satisfied by all $x$ !
So you may apply transformations, but validate the solutions using the original equation.

In your example, you establish
$$x^2+x+1=0\implies x^3-1=0.$$
But as $$x^3-1=0=(x-1)(x^2+x+1),$$ nothing is wrong if you ignore the first factor.
