Why isn't this approach in solving $x^2+x+1=0$ valid?

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.


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 $x^2 + x + 1 = 0$.


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 $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

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Algebra Precalculus

Proof Writing

Proof Verification

Proof Explanation

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