How to prove $(c - b) ^ 2 + 3cb = x^3$ has no nonzero integer solutions? I'm trying solve: 

$a^3 + b^3 = c^3$ has no nonzero integer solutions. 

Only one problem left:  because
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3,\quad (1)$  
if $~c-b~$ is a cubic number, divide both side of $~(1)~$ by $~(c - b)~$ get 
$(c - b) ^ 2 + 3cb = x^3.\quad (2)$  
How to prove equation $~(2)~$ has no nonzero integer solutions for $~c,\ b,\ x$?
Edit: if $~c,\ b,\ x~$ are integers, we can assume they are coprime.

Fermat's theorem cannot be proved, if you do not know how to solve Diophantine equations. 
This approach cannot be used. This equation has a solution.  $$(c-b)^2+3cb=x^3$$ 
The solutions have the form: 
$$x=3(3p^2+s^2)^2$$  
$$c=(3p^2+s^2)(36p^2s^2+36ps(3p^2-s^2)-3(3p^2-s^2)^2)$$ 
$$b=(3p^2+s^2)(36p^2s^2-36ps(3p^2-s^2)-3(3p^2-s^2)^2)$$ 
$p,s$ - any integer.
You can write another solution. This is equivalent to the equation:
$$x^2+xy+y^2=z^3$$
The solutions have the form:
$$x=s^3+3ps^2-p^3$$
$$y=p^3+3sp^2-s^3$$
$$z=p^2+ps+s^2$$
$$...$$
$$x=s(p^2+ps+s^2)$$
$$y=p(p^2+ps+s^2)$$
$$z=p^2+ps+s^2$$
$$...$$
Will make a replacement.
$$b=3p^2+6ps+2s^2$$
$$t=6s^2+6ps$$
$$q=3p^2+6ps+4s^2$$
The solution has the form:
$$x=q(3b^2-6bt-t^2)$$
$$y=q(3b^2+6bt-t^2)$$
$$z=3q^2$$
