<s>[INST] Is there any other method to find the remainder when $x+x^3+x^5+\cdots+x^{2n-1}$ is divided by $x^3-x$? A method I got from somewhere.
\begin{align}
f(x)
&=x+x^3+x^5+\cdots+x^{2n-1}\\
&= x\Bigg[\Big(x^{2(n-1)}-1\Big)+\Big(x^{2(n-2)}-1\Big)+\cdots+\Big(x^{2\times 1}-1\Big)+\Big(x^{2\times 0}-1\Big)+n\Bigg]\\
&=   x\Big(x^{2(n-1)}-1\Big)+x\Big(x^{2(n-2)}-1\Big)+\cdots+x\Big(x^{2\times 1}-1\Big)+x\Big(x^{2\times 0}-1\Big)+nx
\end{align}
As the last term $nx$ cannot be divided by $x^3-x$, the remainder is $nx$.
My Attempt
I attempted to find another method as follows.
\begin{align}
f(x)
&=x+x^3+x^5+\cdots+x^{2n-1}\\
&= x\Big[x^{2\times 0}+x^{2\times 1}+x^{2\times 2}+\cdots+x^{2(n-1)}\Big]\\
&= x\frac{1\times(x^{2n}-1)}{x^2-1}\\
&=\frac{x(x^n-1)(x^n+1)}{(x-1)(x+1)}
\end{align}
From this, how can we find the remainder $nx$ when $\displaystyle\frac{x(x^n-1)(x^n+1)}{(x-1)(x+1)}$  is divided by $x^3-x$?
 [/INST] First, we can eliminate a factor of $x$ , and find the remainder of
$P(x) = 1+x^2+x^4+\cdots+x^{2n-2}$ divided by $x^2-1$
Write $P(x)$ as
$P(x) = (x^2-1)Q(x) + ax + b$
Now substitute the values $x=1$, $x=-1$ to obtain
$P(1) = a+b$
$P(-1) = b-a$
and note that $P(1)=P(-1) = n $, and solve the simultaneous equations...
</s>