6. Let $P_{1}, P_{2}, \ldots, P_{6}$ be points in the complex plane, which are also roots of the equation $x^{6}+6 x^{3}-216=0$. Given that $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6}$ is a convex hexagon, determine the area of this hexagon.
Answer: $9 \sqrt{3}$
Factor $x^{6}+6 x^{3}-216=\left(x^{3}-12\right)\left(x^{3}+18\right)$. This gives us 6 points equally spaced in terms of their angles from the origin, alternating in magnitude between $\sqrt[3]{12}$ and $\sqrt[3]{18}$. This means our hexagon is composed of 6 triangles, each with sides of length $\sqrt[3]{12}$ and $\sqrt[3]{18}$ and with a 60 degree angle in between them. This yields the area of each triangle as $\frac{3 \sqrt{3}}{2}$, so the total area of the hexagon is $9 \sqrt{3}$.