smc

Plotting points $B$ and $C$ on the graph shows that they are at $\left(-x,x^2\right)$ and $\left(x,x^2\right)$, which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}\ast 2x\ast x^2=64=x^3$ Making x=4, and the length of $BC$ is $2x$, so the answer is $\boxed{8}$.