Let $I=\displaystyle\int_{0}^{\pi}x^{3}\sin x\,dx$.

Using integration by parts: let $u=x^{3},\; dv=\sin x\,dx$ $\Rightarrow du=3x^{2}dx,\; v=-\cos x$

$I = [-x^{3}\cos x]_0^{\pi} + \displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx$

Now, let $J=\displaystyle\int_{0}^{\pi}x^{2}\cos x\,dx$

Again, by parts: $u=x^{2},\; dv=\cos x\,dx \Rightarrow du=2x\,dx,\; v=\sin x$

$J = [x^{2}\sin x]_0^{\pi} - \displaystyle\int_{0}^{\pi}2x\sin x\,dx$

Let $K=\displaystyle\int_{0}^{\pi}x\sin x\,dx$ By parts: $u=x,\; dv=\sin x\,dx \Rightarrow du=dx,\; v=-\cos x$

$K=[-x\cos x]_0^{\pi}+\displaystyle\int_{0}^{\pi}\cos x\,dx = \pi$

So $J=0-2K=-2\pi$

Then $\displaystyle\int_{0}^{\pi}3x^{2}\cos x\,dx = 3J = -6\pi$

$I = [-x^{3}\cos x]_0^{\pi} + 3J = (-\pi^{3}\cos\pi - 0) - 6\pi = \pi^{3} - 6\pi$

Answer: $\boxed{\pi^{3} - 6\pi}$ ✅