Find all the triples of positive integers $(a, b, c)$ for which the number

$$
\frac{(a+b)^{4}}{c}+\frac{(b+c)^{4}}{a}+\frac{(c+a)^{4}}{b}
$$

is an integer and $a+b+c$ is a prime.
$(1,1,1),(1,2,2),(2,3,6)$.

Let $p=a+b+c$, then $a+b=p-c, b+c=p-a, c+a=p-b$ and

$$
\frac{(p-c)^{4}}{c}+\frac{(p-a)^{4}}{a}+\frac{(p-b)^{4}}{b}
$$

is a non-negative integer. By expanding brackets we obtain that the number $p^{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ is integer, too. But the numbers $a, b, c$ are not divisible by $p$, therefore the number $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is (non negative) integer. That is possible for the triples $(1,1,1),(1,2,2),(2,3,6)$ only.