<s>[INST] show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
 [/INST] In this answer, the more general integral
$$
\int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x
$$
is calculated.
Your integral is that integral for $n=3$.

A Different Way
In a fashion similar to this answer, we will use the equation
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3}
=\frac{9\sin(3kx)-3\sin(kx)}{4k}\tag{1}
$$
and the series for $0\lt x\le\pi$,
$$
\sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}{2}\tag{2}
$$
Using $(2)$, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{9\sin(3kx)-3\sin(kx)}{4k}
&=\frac94\frac{\pi-3x}{2}-\frac34\frac{\pi-x}{2}\\
&=\frac{3\pi}{4}-3x\tag{3}
\end{align}
$$
Integrating from $0$ twice to back out the derivatives taken in $(1)$ yields
$$
\sum_{k=1}^\infty\frac{\sin^3(kx)}{k^3}=\frac{3\pi}{8}x^2-\frac12x^3\tag{4}
$$
Set $x=1/n$ and multiply by $n^2$ to get
$$
\sum_{k=1}^\infty\frac{\sin^3(k/n)}{k^3/n^3}\frac1n=\frac{3\pi}{8}-\frac1{2n}\tag{5}
$$
and $(5)$ is a Riemann sum for
$$
\int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6}
$$
</s>