\[ I = \int_{0}^{\pi} x^3 \sin x \,dx \] Applying the integration by parts rule, we get \[ = x^3 \int_{0}^{\pi} \sin x \,dx - \int_{0}^{\pi} 3x^2 (-\cos x) \,dx \] \[ = \left[ x^3 (-\cos x) \right]_{0}^{\pi} + 3 \int_{0}^{\pi} x^2 \cos x \,dx - \int_{0}^{\pi} 2x (\sin x) \,dx \] \[ = \pi^3 + 0 - 6 \int_{0}^{\pi} x (\sin x) \,dx \] \[ = \pi^3 - 6 \left[ x \int_{0}^{\pi} \sin x \,dx - \int_{0}^{\pi} (-\cos x) \,dx \right] \] \[ = \pi^3 - 6 \left[ \pi - 0 \right] \] \[ = \pi^3 - 6\pi \] Hence, option (1) is correct.