Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.

Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f(-1)$.
Now since $x^{3}-x = x(x^{2}-1)=x(x-1)(x+1)$ is there any relation between the remainder of $f(x)$ divided by $x^{3}-x$ and remainder when $f(x)$ divided by $(x-1)$ and $(x+1)$?

As the degree of the divisor is $3$, so the degree of the remainder is at most $2$. Let the remainder $R\left(x\right)$ be $ax^2+bx+c$. As $x^3-x=x\left(x-1\right)\left(x+1\right)$, we'll find that: $$\begin{cases} R\left(-1\right)=a-b+c=\left(-1\right)^{81}+\left(-1\right)^{49}+\left(-1\right)^{25}+\left(-1\right)^9+\left(-1\right)=-5 \\ R\left(0\right)=c=0^{81}+0^{49}+0^{25}+0^9+0=0 \\ R\left(1\right)=a+b+c=1^{81}+1^{49}+1^{25}+1^9+1=5 \end{cases}$$ After some calculation, you'll get $a=0,b=5,c=0$, so the remainder is $\boxed{5x}$.
