# X^3 -x -x^3=0 puzzle

1. Feb 10, 2005

### regor60

This a puzzle for me.

x^3 -x -x^3=0 Obviously x=0
but, divide both sides by x

x^2 -1 -x^2=0 except for when x=0 'cause undefined
but that means
-1=0

is this apparent conflict not in fact one because you can't rule out x=0 ?

2. Feb 10, 2005

### Davorak

3. Feb 10, 2005

### The Bob

$$x^3 - x - x^3 = 0$$ then divide by x:

$$\frac{x^3 - x - x^3}{x} = \frac{0}{x}$$

$$x^2 - x^2 = 1$$

Even saying that $$\frac{0}{x}$$ is 0 you should see that it is not possible to get two different numbers to make 1 when the two x values will give the same to take away.

4. Feb 10, 2005

### K.J.Healey

I thought that -1=0 is not a statement of equality, but more of a qualifier for the original equation. We want to know when say, x^2 - x^2 == 1, and that will only happen when 0 = 1, which is never. So that has no solution. I thought this meaning was always implied in mathematics. I mean I can say
1 + 1 =5, but how can that be?! I mean I just wrote it, but its not supposed to be possible! Just becasue yo ucan write down an equation doesnt mean its true.

5. Feb 11, 2005

### gerben

You are simply breaking the rule of "not dividing by zero".
You could state your "problem" simpler:
x = 0
divide both sides by x
x/x = 0/x <-> 1 = 0