First, $u=|z^3-3z+2| = |(z-1)^2(z+2)|$. Let $z=x+yi$ where $x, y \in \mathbb{R}$. We then have $x^2 + y^2 =1$ and $|x|\le 1$. $$u=\sqrt{[(x-1)^2 + y^2]^2[(x+2)^2+y^2]}=\sqrt{(-2x+2)^2(4x+5)}\le\sqrt{\Big[\frac{(2-2x)+(2-2x)+(4x+5)}{3}\Big]^3}=3\sqrt{3}$$