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.
On
Analyzing the binomial expansion sounds like a good idea. Most terms have $abc$ in them, so those are easy. For the ones that don't, however, you can show that each and every one of them is divisible by $abc$.
For instance, $a^2b^{11}$ is divisible by $abc$ because it is equal to $a\cdot b\cdot b^3\cdot ab^7$. The first factor is divisible by $a$, the second is divisible by $b$ and the third is divisible by $c$.
If you don't want this to be really long, you'll have to do it systematically and generally in some way, as there are many terms to handle.
On
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
Consider the combined set of distinct prime factors of $abc$ being $p_i$ for $1 \le i \le n$ for some $n \ge 1$. In particular, you have
$$a = \prod_{i=1}^{n} p_i^{e_i}, \text{ with } e_i \ge 0 \tag{1}\label{eq1A}$$
$$b = \prod_{i=1}^{n} p_i^{f_i}, \text{ with } f_i \ge 0 \tag{2}\label{eq2A}$$
$$c = \prod_{i=1}^{n} p_i^{g_i}, \text{ with } g_i \ge 0 \tag{3}\label{eq3A}$$
The stated divisibility properties means that, for each $1 \le i \le n$, you have
$$3e_i \ge f_i \tag{4}\label{eq4A}$$
$$3f_i \ge g_i \tag{5}\label{eq5A}$$
$$3g_i \ge e_i \tag{6}\label{eq6A}$$
For some given $i$, assume $e_i$ is the minimum of $e_i, f_i$ and $g_i$, so $a$, $b$ and $c$ are each divisible by $p_i^{e_i}$. Thus, the value of
$$(a+b+c)^{13} \tag{7}\label{eq7A}$$
would have at least $13e_i$ factors of $p_i$. From \eqref{eq4A}, you have that $f_i \le 3e_i$ and $9e_i \ge 3f_i$. The latter, combined with \eqref{eq5A}, gives $9e_i \ge 3f_i \ge g_i \implies g_i \le 9e_i$. You therefore have the # of factors of $p_i$ in $abc$ is $e_i + f_i + g_i \le e_i + 3e_i + 9e_i = 13e_i$, which means it's less than or equal to the # of factors of $p_i$ of \eqref{eq7A}. You can repeat this procedure for the cases where $f_i$ or $g_i$ is the minimum instead for any given $i$, and then do this for each $1 \le i \le n$, to prove $abc$ divides the result of \eqref{eq7A}.