How to prove $(c - b) ^ 2 + 3cb = x^3$ has no nonzero integer solutions?

Question

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.

Answer 1

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:

$$x=q(3b^2-6bt-t^2)$$

$$y=q(3b^2+6bt-t^2)$$

$$z=3q^2$$

Answer 2

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).

Answer 3

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.