$a,b,c\in\mathbb{N}\:\land\:\frac ab+\frac bc+\frac ca\in\mathbb{N}\Rightarrow abc=n^3,\:n\in\mathbb{N}$ 
If $a$, $b$ and $c$ are natural numbers such that $$\frac ab+\frac bc+\frac ca\in\mathbb{N},$$ prove that $abc$ is a perfect cube.

All I could get was $ab^2+bc^2+ca^2\equiv_{abc}0$.
 A: Theorem $\rm\displaystyle\ \ For\,\ a,b,c,n\in\Bbb Z\!:\ \frac{a}b+\frac{b}c+\frac{c}a =n\:\Rightarrow\: abc= m^3,\ $ for some $\rm\,m\in \Bbb Z$
Proof $\ $ Scaling by $\rm\,abc\,$ yields $\rm\quad  a^2 c + b^2 a + \color{#C00}{c^2 b}\, =\, nabc\qquad\qquad \color{brown}{(E)}$
We prove  $\rm\,\ a = kj^2/d^2,\, b = ik^2/d^2,\, c = ji^2/d^2,\,\ $ so $\rm\,\ {\bf abc = m^3}\:$ for $\rm\: m = ijk/d^2\in\Bbb Z,\: $ 
for $\rm\ i = (b,c),\ j = (a,c),\  k = (a,b),\  d = (a,b,c).\ $ Note $\rm\:d\mid i,j,k\: $ so $\rm\,d^2\!\mid jk,i^2,j^2,k^2.$ 
$$\begin{eqnarray}\rm j^2 k\, 
&=&\ \ \ \rm (a,c)^2 (a,b)\\
&=&\ \ \ \rm (a^2,ac,c^2)(a,b)\\
&=&\,\rm a(a^2,ac,c^2,ab,cb,\ \ \ \color{#C00}{c^2b/a)}\ \ \ so,\ by\ \color{brown}{(E)}/a\\
&=&\,\rm a(a^2,\color{#e70}{ac},c^2, ab,\color{#0A0}{bc},\overbrace{n\color{#0A0}{bc}\!-\!\color{#e70}{ac}\!-\!b^2)}\\
&=&\,\rm a(a^2,ac,c^2,ab,bc,b^2)\\
&=&\,\rm a(a,b,c)^2 =\, ad^2\\
\end{eqnarray}$$
Finally, by symmetry, also  $\rm\,\ ik^2 = bd^2,\,\ ji^2 = cd^2.\quad$ QED
Remark $ $ There is much literature on this equation, e.g. see Dave Rusin's page and OEIS A085705, where you will find mention of the connection with elliptic curves.  Since the above proof uses only gcds, it holds true (mod units) over any gcd-domain, i.e. any domain where every pair of nonzero elements has a gcd. Hence this proof is more general than proofs employing prime factorization. For example, it works in the ring $\,\Bbb I\,$ of all agebraic integers, where no primes exist (since $\rm\, a = \sqrt{a}\sqrt{a}),\:$ but which, nevertheless, is a Bezout domain, so a gcd domain.
A: For any prime $p$ and rational $x$, denote $v_p(x)$ (the $p$-valuation of $x$) the integer such that $x = p^{v_p(x)}a/b$ where $p$ doesn't divide $a$ nor $b$.
It is easy to show that $v_p(n) \ge 0$ if $n$ is an integer; $v_p(xy) = v_p(x)+v_p(y)$; and $v_p(x+y) \ge \min\{v_p(x), v_p(y)\}$, with an equality if those two valuations are distinct.
A number $x$ is a cube if and only if for every prime $p$, $v_p(x)$ is a multiple of $3$.
Let $p$ be any prime. We have to show that $v_p(abc) \equiv 0 \pmod 3$.
First, notice that $v_p(abc) \equiv v_p(\frac a b) - v_p(\frac b c) \pmod 3 (\equiv v_p(\frac b c) - v_p(\frac c a) \equiv v_p(\frac c a) - v_p(\frac a b))$  
So if any two of the three valuations $v_p(\frac a b), v_p(\frac b c), v_p(\frac c a)$ are equal, we deduce that $v_p(abc)$ is a multiple of $3$.
Suppose that they are distinct. Then $0 \le v_p(\frac a b + \frac b c + \frac c a) = \min \{v_p(\frac a b), v_p(\frac b c), v_p(\frac c a)\}$ (because they are distinct).
And so $0 = v_p(1) = v_p(\frac a b) + v_p(\frac b c) + v_p(\frac c a) \ge 0+1+2 = 3 > 0$, which is impossible.
A: Consider polynom $(X-\frac a b)(X- \frac b c)(X - \frac c a)$.
It is equal to $X^3 - u*X^2 + v*X - 1$, where $u = \frac a b + \frac b c + \frac c a$.
and $v = \frac b a + \frac c b + \frac a c$.
for $X = 1$, you get $1 - u + v - 1 = v - u$. So you have $(1-\frac a b)(1- \frac b c)(1- \frac c a) = v-u$.
You can rewrite that: $(b-a)(c-b)(a-c)=c(b^2-a^2)+a(c^2-b^2)+b(a^2-c^2)$
but not sure it leads anywhere.
