To solve \( \left( 4 \text{cis} \frac{\pi}{18} \right)^3 \) using DeMoivre's Theorem, we first note that DeMoivre's theorem states:
\[ (r \text{cis} \theta)^n = r^n \text{cis}(n\theta) \]
In this case:
- \( r = 4 \)
- \( \theta = \frac{\pi}{18} \)
- \( n = 3 \)
Now we apply the theorem:
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Calculate \( r^n \): \[ r^n = 4^3 = 64 \]
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Calculate \( n \theta \): \[ n\theta = 3 \times \frac{\pi}{18} = \frac{\pi}{6} \]
Putting it all together, we have: \[ \left( 4 \text{cis} \frac{\pi}{18} \right)^3 = 64 \text{cis} \frac{\pi}{6} \]
Now we need to express \( 64 \text{cis} \frac{\pi}{6} \) in rectangular form. Recall that \( \text{cis} \theta = \cos \theta + i \sin \theta \). Therefore:
\[ \text{cis} \frac{\pi}{6} = \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \]
We know:
- \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
- \( \sin \frac{\pi}{6} = \frac{1}{2} \)
Thus: \[ \text{cis} \frac{\pi}{6} = \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \]
Now plug this into our expression: \[ 64 \text{cis} \frac{\pi}{6} = 64 \left( \frac{\sqrt{3}}{2} + i \cdot \frac{1}{2} \right) = 64 \cdot \frac{\sqrt{3}}{2} + 64 \cdot i \cdot \frac{1}{2} \] \[ = 32\sqrt{3} + 32i \]
So, the final answer is: \[ 32\sqrt{3} + 32i \]
Therefore, the correct response is:
32 start root 3 end root plus 32 lower i.