Find the remainder obtained on dividing $p(x)=x^3+1$ by $x+1$.

By long division,

$\overset{\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-x+1}{\mathop{x+1\sqrt{\begin{align}
  & {{x}^{3}}+1 \\ 
 & {{x}^{3}}\pm {{x}^{2}} \\ 
\end{align}}}}\,$

             $\_\_\_\_\_\_\_\_\_\_\_\_\_$

             $-{{x}^{2}}+1$

             $\mp {{x}^{2}}\pm {{x}}$

            $\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

                   $x+1$

                  $- x \pm 1$

             $\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

                       $0$

So, we find that the remainder is $0$.

Here $p(x) = x^3 + 1$, and the root of $x + 1 = 0$ is $x = -1$. We see that

$p(-1) = (-1)^3 + 1$

$= -1 + 1 $

$= 0$