This is a well known problem but it has a long solution... first we assume gcd(a,b,c) to be 1 because otherwise gcd is simplified. Then we assume p to be a prime factor of abc and assume a=a'p Then we substitute into the given expression by putting in the same denominator.. and we go on like this expressing b in the same manner and then at the end we get that p^3|abc and as p is random, it means every prime divisor of abc has the power 3 so abc is a perfect cube look at : http://answers.yahoo.com/question/index?qid=20100124161119AAfBOXz

Suppose that $\sum_{cyc}\frac{a}{b}=n$. WLOG, Suppose $(a,b,c)=1$. Take any $p \in \mathbb{P}$ such that $p|abc$. Now, if $p|a$, since $a^2c+b^2a+c^2b=nabc \implies p|bc^2 \implies p|b$ or $p|c$. Similarly, any prime divisor of $a$,$b$ or $c$ must divide exactly $2$ of them. Take a $p \in \mathbb{P}$ such that $p|a,p|b$. (All other possible cases are totally similar). If $v_p(b) < 2v_p(a)$ $\implies v_p(b)+1 \le 2v_p(a)$ $\implies p^{v_p(b)+1}|a^2c,b^2a,nabc$ $\implies p^{v_p(b)+1}|c^2b \implies p|c$, a contradiction. If $v_p(b) > 2v_p(a)$ $\implies v_p(b) \ge 2v_p(a)+1$ $\implies p^{2v_p(a)+1}|b^2a, c^2b,nabc$ $\implies p^{2v_p(a)+1}|a^2c \implies p|c$, a contradiction. So, $v_p(b)=2v_p(a) \implies abc= \prod_{p|abc}p^{3v_p(a)}$ which is a perfect cube.

We have $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} = \frac{ab^2+bc^2+ca^2}{abc}$, so $abc \mid ab^2+bc^2+ca^2$. Let $p$ be a prime, and write $a = p^e a_0$, $b = p^f b_0$, and $c = p^g c_0$, where $e, f, g \ge 0$ and $p \nmid a_0, b_0, c_0$. Without loss of generality, let $e \le f \le g$. Substituting, we have \[p^{e+f+g}a_0b_0c_0 \mid p^{e+2f}a_0b_0^2+p^{f+2g}b_0c_0^2+p^{g+2e}c_0a_0^2.\]Since $e + f + g \le f + 2g$, we may remove the term with $p^{f+2g}$, giving \[p^{e+f+g}a_0b_0c_0 \mid p^{e+2f}a_0b_0^2 + p^{g+2e}c_0a_0^2.\]Dividing both sides by $p^{2e+f}$, we obtain \[p^{g-e}a_0b_0c_0 \mid p^{f-e}a_0b_0^2 +p^{g-f}c_0a_0^2,\]so \[p^{g-e} \mid p^{f-e}a_0b_0^2 + p^{g-f}c_0a_0^2,\]or \[p^{m+n} \mid p^m a_0b_0^2 + p^n c_0a_0^2\]where $m = f-e$ and $n = g-f$. If $m > n$, then \[p^{m+n} \mid p^ma_0b_0^2 + p^nc_0a_0^2 = p^n(p^{m-n}a_0b_0^2+c_0a_0^2),\]so $p^m \mid p^{m-n} a_0b_0^2 + c_0a_0^2$. But since $p \nmid a_0, b_0, c_0$ we have $p \nmid p^{m-n} a_0b_0^2 + c_0a_0^2$, a contradiction. Similarly, if $m < n$, then \[p^{m+n} \mid p^ma_0b_0^2 + p^nc_0a_0^2 = p^m(a_0b_0^2+p^{n-m}c_0a_0^2),\]but $p \nmid a_0b_0^2 + p^{n-m}c_0a_0^2$, a contradiction. Hence $m = n$, so $f-e = g-f$, and the sequence $e, f, g$ is arithmetic. Then $p^e \cdot p^f \cdot p^g = p^{e+f+g} = p^{3f}$, which is a perfect cube. Since $p$ was chosen arbitrarily, this is enough to show that $abc$ is a perfect cube.

At above it does as $p|a$!

Ah, I didn't realize they specified $(a,b,c)=1$, oops.