<s>[INST] $a,b,c\in\mathbb{N}\:\land\:\frac ab+\frac bc+\frac ca\in\mathbb{N}\Rightarrow abc=n^3,\:n\in\mathbb{N}$ 
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 $ab^2+bc^2+ca^2\equiv_{abc}0$.
 [/INST] 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.
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