In the complex plane, let $A$ be the set of solutions to $z^3-27=0$ and let $B$ be the set of solutions to $z^3-12z^2-12z+144=0$. What is the greatest distance between a point of $A$ and a point of $B?$

$A=\{3,-\frac32+\frac{3\sqrt3}2i,-\frac32-\frac{3\sqrt3}2i\}$ (cube roots of $27$). Factoring $z^3-12z^2-12z+144=(z-12)(z^2-12)$ gives $B=\{12,2\sqrt3,-2\sqrt3\}$.

All points of $A$ lie on the circle $|z|=3$, so a point $b\in B$ is at most $|b|+3$ from any point of $A$. For $b=\pm2\sqrt3$ this is at most $2\sqrt3+3<7$, while $12$ is $9$ away from $3$ and even farther from the other two cube roots. So the greatest distance is between $12$ and $-\frac32+\frac{3\sqrt3}2i$ (or its conjugate):
$$\left|\left(-\frac32-12\right)+\frac{3\sqrt3}2i\right|=\sqrt{\left(\frac{27}2\right)^2+\left(\frac{3\sqrt3}2\right)^2}=\sqrt{\frac{729+27}4}=\sqrt{189}=\boxed{3\sqrt{21}}.$$