It is given that
\(\begin{array}{l}
x+\frac{1}{x}=2 \sin \alpha \Rightarrow x=\sin \alpha+i \cos \alpha=\cos \left(\frac{\pi}{2}-\alpha\right) +i \sin \left(\frac{\pi}{2}-\alpha\right) \\
\Rightarrow \quad x=e^{i\left(\frac{\pi}{2}-\alpha\right)}
\end{array}\)
and \(y+\frac{1}{y}=2 \cos \beta \Rightarrow y=\cos \beta+i \sin \beta=e^{i \beta}\)
Now, \(x^3 y^3=e^{i\left(\frac{3 \pi}{2}-3 \alpha\right)} \cdot e^{i 3 \beta}=e^{i\left(\frac{3 \pi}{2}+(3 \beta-3 \alpha)\right)}\)
\(\begin{aligned}
& =\cos \left(\frac{3 \pi}{2}+3 \beta-3 \alpha\right)+i \sin \left(\frac{3 \pi}{2}+3 \beta-3 \alpha\right) \\
& =\sin (3 \beta-3 \alpha)-i \cos (3 \beta-3 \alpha)
\end{aligned}\)
and \(\frac{1}{x^3 y^3}=\sin (3 \beta-3 \alpha)+i \cos (3 \beta-3 \alpha)\)
So, \(x^3 y^3+\frac{1}{x^3 y^3}=2 \sin (3 \beta-3 \alpha)=2 \sin 3(\beta-\alpha)\)
Hence, option (c) is correct.