jmc

We notice that $x^3-\frac{1}{x^3}$ is a difference of cubes. We can therefore factor it and rearrange the terms to get: $$\begin{array}{rl}{x}^3-\frac{1}{x^3}& =\left(x-\frac{1}{x}\right)\cdot \left(x^2+x\left(\frac{1}{x}\right)+\frac{1}{x^2}\right)\\ & =\left(x-\frac{1}{x}\right)\cdot \left(\left(x^2-2x\left(\frac{1}{x}\right)+\frac{1}{x^2}\right)+3x\left(\frac{1}{x}\right)\right)\\ & =\left(x-\frac{1}{x}\right)\cdot \left({\left(x-\frac{1}{x}\right)}^2+3\right).\end{array}$$Since $x-\frac{1}{x}=4$, we have that $x^3-\frac{1}{x^3}=4\cdot (4^2+3)=4\cdot 19=\boxed{76}.$