$x^{10}+x^{11}+\dots+x^{20}$ 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)


What about this : We have for example $$x^{10}+x^{12}=x^{10}(x^2+1)\equiv 0\mod x(x^2+1)$$
This way we can also cancel $11-13,14-16,15-17,18-20$. It remains $x^{19}$ for which you can use $x^6\equiv x^2$ giving $x^{18}\equiv x^6$ hence $x^{18}\equiv x^2$ hence $x^{19}\equiv x^3\equiv -x$ $\mod (x^3+x)$
