Given $a^3 + b^{3}+ c^{3}= (a+b+c)^{3} $.
Prove that for any natural number $n$, $$a^{2n+1}+b^{2n+1}+c^{2n+1}=(a+b+c)^{2n+1}$$
I first tried mathematical induction but did not proceed anywhere.
Can the formula $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)$ be directly used?
Can this problem be solved using mathematical induction?
Solution 1 Using Induction
We have $$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)\implies(a+b)(b+c)(c+a)=0$$
Now, for $k$ we have
$$(a+b+c)^{2k+1}=a^{2k+1}+b^{2k+1}+c^{2k+1}$$ For $(k+1)$ $$(a+b+c)^{2k+3}=(a+b+c)^2(a+b+c)^{2k+1}=(a+b+c)^2(a^{2k+1}+b^{2k+1}+c^{2k+1})$$
$$(a+b+c)^{2k+3}=a^{2k+3}+b^{2k+3}+c^{2k+3}+(a+b)(b+c)(c+a)(\text{some factor})$$
(I've left the job of finding factor to you for observing pattern see here and here)
Using $(a+b)(b+c)(c+a)=0$ $$(a+b+c)^{2k+3}=a^{2k+3}+b^{2k+3}+c^{2k+3}$$
Solution 2
$$(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3(a+b)(b+c)(c+a)\implies(a+b)(b+c)(c+a)=0$$ Either of $a+b,b+c,c+a =0$
$\implies (a+b+c)=a$ or $(a+b+c)=b$ or $(a+b+c)=c$
In any case $$(a+b+c)^{2n+1}=a^{2n+1}+b^{2n+1}+c^{2n+1}$$
We have $$b^3+c^3=(a+b+c)^3-a^3$$
$$\iff(b+c)(b^2-bc+c^2)=(a+b+c-a)\{(a+b+c)^2+a(a+b+c)+a^2\}$$
$$\iff(b+c)\{3a^2+3ab+3bc+3ca\}=0$$
$$\iff(b+c)(a+b)(a+c)=0$$
As the product of three terms is zero, hence at least one of them must be zero
If $b+c=0, a+b+c=a\implies (a+b+c)^{2n+1}=?$
and using Prove by induction that $a-b|a^n-b^n$,
$b^{2n+1}+c^{2n+1}=b^{2n+1}-(-c)^{2n+1}$ is diviisble by $b-(-c)=b+c$
$\implies b+c=0\implies b^{2n+1}+c^{2n+1}=0\implies a^{2n+1}+b^{2n+1}+c^{2n+1}=? $
Similarly, if $c+a=0$ or if $a+b=0$
