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.
 A: 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:
$$x=q(3b^2-6bt-t^2)$$
$$y=q(3b^2+6bt-t^2)$$
$$z=3q^2$$
A: Let me have a try:
Suppose $~c,\ b,\ x~$ are nonzero integers, We can assume that all variables are coprime,let $~(c-b)^2=y^3,~$ from $~(2)~$ get:
$x^3-y^3=(x-y)((x-y)^{2}+3xy)=3cb\quad (3)$
if $~x-y=3~$ divide both side of  $(3)\ $ by $~3$:
$9 + 3xy= cb,\quad (4)$ 
form $~(4)~$ we see $~(xy,\ cb)>1~$,this conflict by assuming, if $~x-y \neq 3~$ ,by the same way, we can also get some conflict  (lack of detail).
A: This is not an answer for your question. I wont to show that some times this kind of Diophantine equations may have solutions in positive integers.
Consider the equation $$ x^2+y^2=z^3.$$ If we take $$x=n(n^2-3)$$ $$y=3n^2-1$$ for $n>1,$  then the above equation has infinitely many integer solutions.
