8. \( \left[\sqrt{3}\left(\cos 15^{\circ}+i \operatorname{sen} 15^{\circ}\right)\right]^{9} \)

To simplify the expression \( \left[\sqrt{3}\left(\cos 15^{\circ}+i \sin 15^{\circ}\right)\right]^{9} \), we can use the polar form of complex numbers and De Moivre's theorem. First, we recognize that \( \cos 15^{\circ} + i \sin 15^{\circ} \) can be expressed as \( e^{i 15^{\circ}} \). Therefore, we can rewrite the expression as: \[ \sqrt{3} e^{i 15^{\circ}} \] Now, we raise this to the 9th power: \[ \left[\sqrt{3} e^{i 15^{\circ}}\right]^{9} = (\sqrt{3})^{9} \cdot (e^{i 15^{\circ}})^{9} \] Calculating \( (\sqrt{3})^{9} \): \[ (\sqrt{3})^{9} = (3^{1/2})^{9} = 3^{9/2} = 3^{4.5} = 3^4 \cdot 3^{0.5} = 81 \sqrt{3} \] Next, we calculate \( (e^{i 15^{\circ}})^{9} \): \[ (e^{i 15^{\circ}})^{9} = e^{i (15^{\circ} \cdot 9)} = e^{i 135^{\circ}} \] Now we can combine these results: \[ \left[\sqrt{3}\left(\cos 15^{\circ}+i \sin 15^{\circ}\right)\right]^{9} = 81 \sqrt{3} \cdot e^{i 135^{\circ}} \] Next, we convert \( e^{i 135^{\circ}} \) back to rectangular form: \[ e^{i 135^{\circ}} = \cos 135^{\circ} + i \sin 135^{\circ} \] Calculating \( \cos 135^{\circ} \) and \( \sin 135^{\circ} \): \[ \cos 135^{\circ} = -\frac{1}{\sqrt{2}}, \quad \sin 135^{\circ} = \frac{1}{\sqrt{2}} \] Thus, \[ e^{i 135^{\circ}} = -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \] Now we multiply this by \( 81 \sqrt{3} \): \[ 81 \sqrt{3} \left(-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right) = 81 \sqrt{3} \left(-\frac{1}{\sqrt{2}}\right) + 81 \sqrt{3} \left(i \frac{1}{\sqrt{2}}\right) \] This simplifies to: \[ -\frac{81 \sqrt{3}}{\sqrt{2}} + i \frac{81 \sqrt{3}}{\sqrt{2}} \] Thus, the final result is: \[ \boxed{-\frac{81 \sqrt{3}}{\sqrt{2}} + i \frac{81 \sqrt{3}}{\sqrt{2}}} \]