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

Just as Karvens comments: 
Let $a=x-y$ and $b=y-z$. Then $-(a+b)=z-x$. Clearly, $a,b,c \in \Bbb Z$. So 
$$30=a^3+b^3-(a+b)^3=(a+b)(a^2-ab+b^2)-(a+b)^3=-3ab(a+b)$$
And hence 
$$10=-ab(a+b).$$
Therefore $a=\pm 1$ or $a=\pm 2$ or $a=\pm 5$ or $a=\pm 10$. Claerly they cannot be the solution. 
