Note that, $(a+b+c,a)=(a+b+c,b+c)=1$. Moreover, we also have, $(a+b+c,a+b)=(a+b+c,c)=1$ and $(a+b+c,a+c)=(a+b+c,b)=1$. Therefore, $(a+b)(a+c)(b+c)$ and $a+b+c$ are coprime. Next, let $p\mid a+b$ be a prime number. We shall prove that $p\nmid ab+bc+ca$. Assume the converse. Let $p\mid ab+bc+ca$. Then $p\mid a+b\Rightarrow p\mid ac+bc$, and thus, $p\mid ab$. Thus, either $p\mid a$, in which case $p\mid a+b$ yields $p\mid b$, contradicting with the coprimality of $a$ and $b$. Similar holds for $p\mid b$. Thus, ${\rm gcd}((a+b)(a+c)(b+c),ab+bc+ca)=1$. Now, this yields that, ${\rm gcd}((a+b)(a+c)(b+c),(a+b+c)(ab+bc+ca))=1$. Since the product is a perfect power, it therefore holds that for some $m,k$ integers, $(a+b)(a+c)(b+c)=m^n$ and $(a+b+c)(ab+bc+ca)=k^n$. Thus, $$ abc= (a+b+c)(ab+bc+ca)-(a+b)(a+c)(b+c)=k^n-m^n, $$as claimed.