Prove by contradiction that $(x-y)^3+(y-z)^3+(z-x)^3 = 30$ has no integer solutions By factorizing it I found that 
$(x-y)(y-z)(z-x) = 10$

Set $a=x-y$, $b=y-z$ and $c=z-x$. Then $a+b+c=0$, and hence, as $abc=10$ two are negative and one positive. By symmetry, assume
$$
A=-(x-y)>0, \,\, B=-(y-z)>0, \,\, A+B=z-x>0,
$$ 
and for the integers $A,B,C$ we have
$$
(A+B)^3-A^3-B^3=30.
$$
Now as $AB(A+B)=10$, then, as $A+B>A,B$, 
$$
(A,B,A+B)\in \{(1,1,10),(1,2,5), (2,1,5)\},
$$
but none of the triplets works.
Hence no integer solution.
