algebra precalculus - Problem on multiplication formulae.

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?

Answer

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 $$(a+b+c)^{2n+1}=a^{2n+1}+b^{2n+1}+c^{2n+1}$$