What is wrong with the proof?

Algebra Level 3

Let $x$ be a solution of $x^{2} + x + 1 = 0$
Since $x$ $x$ is not 0, we can divide both sides by $x$ $x$ :
- $x + 1 + \frac{1}{x} = 0$
Substitute $x + 1 = -x^{2}$ $x + 1 = - x^{2}$
- $\frac{1}{x} = x^{2}$
$x = 1$ $x = 1$
- $x = x^{3}$
$3 = 0$ $3 = 0$
- $1^{2} + 1 + 1 = 0$

Nothing is wrong! Step 2 because

x

is imaginary Step 1 because there are no solutions Step 3 because it adds an extra solution

3 solutions

X X
May 26, 2018

The beginning equation has only 2 solutions, $\omega,\omega^2$ .But there are 3 solutions for $\frac1x=x^2,x=\omega,\omega^2,1$

The 2 solutions are $w$ and $w^2$ (not $-w$ ); right?

Zeeshan Ali - 3 years ago

Oops,you're right,I mistyped.I editted,thanks for pointing out.

X X - 3 years ago

Greylyn Gao
May 25, 2018

Step 3 because it adds an extra solution due to increasing the degree, then the extra solution is subbed back into the original equation.

Edwin Gray
May 28, 2018

x = 1 is not a root of the original equation, and therefore is extraneous. We did not start with x^3 -1 = (x -1)(x^2 + x + 1)=0. Ed Gray