jmc

By AM-GM, $$\begin{array}{rl}{x}^2+8x+\frac{64}{x^3}& =x^2+2x+2x+2x+2x+\frac{32}{x^3}+\frac{32}{x^3}\\ & \ge 7\sqrt[7]{(x^2)(2x)(2x)(2x)(2x)\left(\frac{32}{x^3}\right)\left(\frac{32}{x^3}\right)}\\ & =28.\end{array}$$Equality occurs when $x=2,$ so the minimum value of $f(x)$ for $x>0$ is $\boxed{28}.$