We use the following facts: 1:$a|b^4,b|c^4\Rightarrow a|b^4|c^{16}\Rightarrow ab|c^4c^{16}=c^{20}$ Similarly we get $b|a^{20}, c|a^{20}$ 2:Since $c|b^4$ and $c|a^{16}$, the $c|a^x b^y$ for all $x,y$ positive integers with $x+y\geq 19$ Now analize the terms in the expansion of $(a+b+c)^{21}$.We'll show that all the terms are multiples of $abc$. For that notice we only need to worry about the terms that don't have $abc$, in other words, all terms that only have one or two beetween a,b,c. Lets only consider terms that have $a$ and $ab$,the rest being competely similar 1:$abc|a^{21}\iff bc|a^{20}$, which we know is true 2:$abc|a^x b^y\iff c|a^{x-1}b^{y-1}$, which we know is true because x-1+y-1=19 Then $abc$ divides all terms in $(a+b+c)^{21}$.

parmenides51 wrote: Let $a,b,c$ be positive integers such that $a|b^4, b|c^4$ and $c|a^4$. Prove that $abc|(a+b+c)^{21}$ generality: Let $a,b,c$ be positive integers such that $a|b^n, b|c^n$ and $c|a^n$. Prove that $abc|(a+b+c)^{n^2+n+1}$.