Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?

Well $a^3 + b^3 = (a+b)(a^2 -ab + b^2)=1$
The only factors of $1$ are either $1\cdot 1$ or $-1 \cdot (-1)$.
So we must have $a+b = a^2 - ab + b^2 =\pm 1$.
And that just won't work unless $ab=0$.
It would follow that $b= \pm 1 -a$
And therefore  $a^2 - a(\pm 1 - a) + (\pm 1 - a) = \pm 1$ so
$a^2+a^2 \mp a -a \pm 1 = \pm 1$
SO either $2a^2 -2a+1=1$ or $2a^2 -1 = -1$ so
$2a(a-1) = 0;b= 1-a$ or $2a^2 = 0;b = -1-a$.
So we have either $a=0$ and $b=1$.  Of $a =1$ and $b =0$.  Or $a=0$ and $b =  -1$.
In all three cases we must have $ab = 0$.
