jmc

By Factor Theorem, we want $[p(x){]}^3-x$ to be equal to 0 at $x=1,$ $x=-1,$ and $x=8.$ Thus, $p(1)=1,$ $p(-1)=-1,$ and $p(8)=2.$

Since $p(x)$ is quadratic, let $p(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}a+b+c& =1,\\ a-b+c& =-1,\\ 64a+8b+c& =2.\end{array}$$Solving this system, we find $a=-\frac{2}{21},$ $b=1,$ and $c=\frac{2}{21}.$ Hence, $$p(x)=-\frac{2}{21}{x}^2+x+\frac{2}{21},$$so $p(13)=-\frac{2}{21}\cdot 13^2+13+\frac{2}{21}=\boxed{-3}.$