# Fermat's last theorem for case of n=3

I'm aware that one can use a method of infinite descent, or even just refer to the more general case which has been proven by Andrew Wiles, but I was thinking about it the other day, and I remember seeing a proof which used the equation

$$\left(x+y-z\right)^3=x^3+y^3-z^3+3\left(x+y\right)\left(x-z\right)\left(y-z\right)$$

And assuming you had positive integer solutions so that

$$x^3+y^3-z^3=0$$

You eventually get to a contradiction which shows that your prior assumption can never be true.

I had a go at it, and as we know that z must be in the form of

$$x+y-3a$$

With a being some positive integer, you can form the equation.

$$\left(3a\right)^3=3\left(x+y\right)\left(3a-x\right)\left(3a-y\right)$$

And then I believe you should be able to find some contradiction.

I don't think the original proof that I saw used this technique, but I have scoured the internet and found nothing on this method of forming the equation

$$\left(x+y-z\right)^3=x^3+y^3-z^3+3\left(x+y\right)\left(x-z\right)\left(y-z\right)$$

And then finding a contradiciton. I believe (if my memory has served me well) the original proof looked at the fact that x,y and z must all be coprime and then there is a contradiction, however I could just be imagining that.

If anyone has the proof or knows how to do it that would be very useful as this has been bugging me recently.