prove $(a+b+c)^n=a^n+b^n+c^n$ if $(a+b+c)^3=a^3+b^3+c^3$

Question

if $(a+b+c)^3=a^3+b^3+c^3$ and n is odd number,prove that: $$(a+b+c)^n=a^n+b^n+c^n$$ hint of the question was:

factor this expression $f(a,b,c)=(a+b+c)^3-(a^3+b^3+c^3)$

after factorization $(a+b+c)^3-(a^3+b^3+c^3)=3(a+b)(b+c)(c+a)$

and another thing i have done $$f(\sqrt [3] a^n,\sqrt [3]b^n,\sqrt [3]c^n)=(\sqrt [3] a^n+\sqrt [3]b^n+\sqrt [3]c^n)^3-(a^n+b^n+c^n)=3(\sqrt [3]a^n+\sqrt [3]b^n)(\sqrt [3]b^n+\sqrt [3]c^n)(\sqrt [3]c^n+\sqrt [3]a^n)$$

Answer 1

If $(a+b+c)^3=a^3+b^3+c^3$, then $a+b=0$ or $a+c=0$ or $b+c=0$ by your factorization.

Changing the names, we may assume $a+b=0$ (say). Then $(a+b+c)^n=c^n$ and $a^n+b^n=0$ ...

Answer 2

Hint: By hypothesis, $f(a,b,c) = (a+b+c)^3 - (a^3+b^3+c^3)=0$. By your factorisation, this means that either $a+b=0$, $b+c=0$ or $c+a=0$.

Answer 3

Your factorization implies $(a+b+c)^3=a^3+b^3+c^3$ iff $3(a+b)(b+c)(c+a)=0$, i.e., iff some pair of variables are of opposite sign. If $a=-b$, for instance, what does that say about $a^n$ and $b^n$ (for odd $n$)?