TITLE: Prove that $(a + b + c)^{13}$ is divisible by $abc$ if $b|a^3$, $c|b^3$ and $a|c^3$.
QUESTION [11 upvotes]: From the Leningrad Mathematical Olympiads:

The natural numbers $a$, $b$ and $c$ have the property
  that $a^3$ is divisible by $b$, $b^3$ is divisible by $c$ and $c^3$ is divisible by $a$. Prove that $(a + b + c)^{13}$ is
  divisible by $abc$.

I thought about analyzing binomial expansion, but it's not being productive. I didn't find similar question in the search.

REPLY [6 votes]: a genral way of solving: lets notice that all the elements in $(a+b+c)^{13}$ are in the form of $a^ib^jc^k$, s.t $i+j+k=13$. we will show that every one of them is divisible by abc: lets divide to cases:
case 1: $i,j,k\ge 1$, which is simple to show.
case 2: one of $i,j,k$ is equal to zero, without loss generality lets assume $k=0$. so it can be written as $a^i*b^j$ s.t $i+j=13$. sub cases are that or $i\ge 10$ or $j\ge 4$, and in both of them we can $a*a^{i-1}*b*b^{j-1}$, and or $a^{i-1}$ is at least $a^9$ and divisible by $b^3$ and therefore also by c, or $b^{j-1}$ is at least $b^3$ and divisible by c.
case 3: two of $i,j,k$ is equal to zero, without loss generality lets assume $j,k=0$.so it can be written as $a^i$, and actually as $a^{13}$, which can be written $a*a^3*a^9$, which $b|a^3$ and $c|b^3$ and $b^3|a^9$ so also divided by c.
after all, all the elements by themselves is divisible by $abc$ and therefor the expression itself