jmc

We try to express $w^3+z^3$ in terms of $w+z$ and $w^2+z^2.$ We have, by sum of cubes, $$w^3+z^3=(w+z)(w^2+z^2-wz),$$so we want now to express $wz$ in terms of $w+z$ and $w^2+z^2.$ To do that, we write $(w+z{)}^2=w^2+z^2+2wz,$ from which it follows that $wz=\frac{1}{2}\left((w+z{)}^2-(w^2+z^2)\right).$ Thus, $$\begin{array}{rl}{w}^3+z^3& =(w+z)(w^2+z^2-\frac{1}{2}\left((w+z{)}^2-(w^2+z^2)\right))\\ & =(w+z)\left(\frac{3}{2}(w^2+z^2)-\frac{1}{2}(w+z{)}^2\right).\end{array}$$Taking magnitudes of both sides, we have $$\begin{array}{rl}\left|w^3+z^3\right|& =\left|(w+z)\left(\frac{3}{2}(w^2+z^2)-\frac{1}{2}(w+z{)}^2\right)\right|\\ & =|w+z|\cdot \left|\frac{3}{2}(w^2+z^2)-\frac{1}{2}(w+z{)}^2\right|.\end{array}$$We are given that $|w+z|=1,$ so $$|w^3+z^3|=\left|\frac{3}{2}(w^2+z^2)-\frac{1}{2}(w+z{)}^2\right|.$$We have $\left|\frac{3}{2}(w^2+z^2)\right|=\frac{3}{2}\cdot 14=21$ and $\left|\frac{1}{2}(w+z{)}^2\right|=\frac{1}{2}\cdot 1^2=\frac{1}{2},$ so by the triangle inequality, $$|w^3+z^3|\ge \left|21-\frac{1}{2}\right|=\boxed{\frac{41}{2}}.$$