($a^n$+$b^n$) is divisible by ($a$+$b$) when $n$ is odd (proof by induction) I am trying to prove that ($a^n$+$b^n$) is divisible by ($a$+$b$) when $n$ is odd. 
I start off by checking for $n$ = $1$, and find that it is true for $n$ = $1$
Now I assume that it is true for $n$ = ($2k$-$1$), 
i.e [$a^{2k-1}$+$b^{2k-1}$] is divisible by ($a$+$b$) 
Finally, I proceed to checking for $n$ = ($2k$+$1$) : 
$=$ {$a^{2k+1}$+$b^{2k+1}$}
$=$ {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$} 
Now I don’t understand how to prove that {$a^2$$a^{2k-1}$+$b^2$$b^{2k-1}$} is divisible by ($a$+$b$) based on my assumption earlier. 
Appreciate any help

Hint:
$$a^{2k+1}+b^{2k+1}=a^2(a^{2k-1}+b^{2k-1})-b^{2k-1}(a^2-b^2)$$
