Prove that $(a + b + c)^{13}$ is divisible by $abc$ if $b|a^3$, $c|b^3$ and $a|c^3$.
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.
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}.
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