$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:

If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$

In these types of questions generally I follow the following approach:

Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$

But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:

 Option (B)


$\overbrace{\!\!\bmod \color{c00}{x^2\!+\!1}}^{\color{#c00}{\large x^2\ \equiv\ -1}\!\!}\!:\ f\color{#0af}{\!+\!x^9} =\! (x^9\!+\!x^{13}\!+\!x^{17})\!\!\overbrace{(1\!+\!x\!+\!x^2\!+\!x^3)}^{\large  ((\color{#c00}{x^2})^2-1)/(x-1)=\color{#0a0}0}\!\!\equiv\color{#0a0}0\,$ so $\,f \equiv \color{#0af}{-x^9} = -x(\color{#c00}{x^2})^4\! \equiv -x$
