Remainder of trinomial What will be the remainder if I divide $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ by $(a+b+c)^3-a^3-b^3-c^3$.
I have tried trinomial expention. But its still too big for long division. Is there a shorter method for this problem?

Since $$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(a+c)(b+c)$$ and
$$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}=0$$ for $a=-b$, for $a=-c$, for $b=-c$ 
and $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divided by $3$, 
we obtain that the remainder is $0$.
