<s>[INST] If $~a^3 + b^3 = c^3~$ has nonzero integer solutions and $~c-b~$ is a cubic number and $~c-b \neq 1$ If $~a^3 + b^3 = c^3~$ has nonzero integer solutions, because:
$c^3 - b ^ 3 = (c - b)((c - b) ^ 2 + 3cb) = a ^ 3,\quad (1)$  
if $~c-b~$ is a cubic number and $~c-b \neq 1~$, divide both side of $~(1)~$ by $~c - b~$ get 
$(c - b) ^ 2 + 3cb = x^3.\quad (2)$ 
It seems $~3 \mid \ x~$ and  $~3 \mid \ c~$ or $~b,~$ how to prove this?
Edit: add tag number-theory 
 [/INST] Let  $~(c−b)^2=y^3,~$ from  $~(2)~$ get: 
$x^3−y^3=(x−y)((x−y)^2+3xy)=3cb,\quad (3) $ 
if $~x-y \neq 3,c,b~$ suppose $~(x-y) \mid c,~$ divide both side of  $(3)\ $ by $~3$:
$(x−y)^2+3xy=3c_1b,\quad (4) $ 
form $~(4)~$ we see $~3 \mid (x−y)^2,~$ because  $~(x-y) \neq 3,~$then  $~3 \mid c_1,~$
form $~(4)~$ get $~3^2k^2+3xy=9c_2b,\quad (5) $ 
divide both side of  $(5)\ $ by $~3$:
$~3k^2+xy=3c_2b,\quad (6) $ 
form $~(6)~$ we see $~3 \mid x~$ or $~y,~$ if $~3 \mid x,~$ problem solved, if $~3 \mid y,~$ form $~(2)~$ we see $~3 \mid x.$ 
I dont know is the proof above right and the cases when $~(x-y)=3~$or $~c~$ or $~b.$
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