1. **State the problem:** We want to simplify the expression $$\sqrt{2} \left(\cos \frac{5\pi}{24} + i \sin \frac{5\pi}{24}\right)^6$$. 2. **Recall the formula:** For a complex number in polar form $$r(\cos \theta + i \sin \theta)$$, raising it to the power $$n$$ is given by De Moivre's theorem: $$\left[r(\cos \theta + i \sin \theta)\right]^n = r^n \left(\cos n\theta + i \sin n\theta\right)$$. 3. **Apply the formula:** Here, $$r = \sqrt{2}$$, $$\theta = \frac{5\pi}{24}$$, and $$n = 6$$. Calculate $$r^n$$: $$r^6 = \left(\sqrt{2}\right)^6 = \left(2^{1/2}\right)^6 = 2^{3} = 8$$. Calculate the new angle: $$6 \times \frac{5\pi}{24} = \frac{30\pi}{24} = \frac{5\pi}{4}$$. 4. **Write the simplified form:** $$8 \left(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}\right)$$. 5. **Evaluate the trigonometric functions:** $$\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$ and $$\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$$. 6. **Substitute these values:** $$8 \left(-\frac{\sqrt{2}}{2} + i \left(-\frac{\sqrt{2}}{2}\right)\right) = 8 \left(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right)$$. 7. **Simplify the expression:** $$8 \times -\frac{\sqrt{2}}{2} = -4\sqrt{2}$$. So the final answer is: $$-4\sqrt{2} - 4\sqrt{2}i$$. This is the rectangular form of the complex number after raising to the 6th power.