To solve $(-1 - i)^9 + (-4 + 2i)^6,$ we can break it into two parts and evaluate each complex number raised to the respective powers.

Part 1: $(-1 - i)^9$

1. Convert to polar form: Let $z = -1 - i$.

   • The modulus $|z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$.
   • The argument $\theta = \tan^{-1} \left( \frac{-1}{-1} \right) = -\frac{3\pi}{4}$.

   So, in polar form: $z = \sqrt{2} \text{cis} \left( -\frac{3\pi}{4} \right),$ where $\text{cis}(\theta) = \cos(\theta) + i \sin(\theta)$.

2. Apply De Moivre's Theorem: [ (-1 - i)^9 = \left( \sqrt{2} \right)^9 \text{