jmc

We can write $$\frac{x^2}{x-8}=\frac{x^2-64+64}{x-8}=\frac{(x-8)(x+8)+64}{x-8}=x+8+\frac{64}{x-8}=x-8+\frac{64}{x-8}+16.$$By AM-GM, $$x-8+\frac{64}{x-8}\ge 2\sqrt{(x-8)\cdot \frac{64}{x-8}}=16,$$so $$\frac{x^2}{x-8}\ge 32.$$Equality occurs when $x=16,$ so the minimum value is $\boxed{32}.$