We know that $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

If $x+y+z=0,$ then $x^{3}+y^{3}+z^{3}-3 x y z=0$ or $x^{3}+y^{3}+z^{3}=3 x y z$

We have to find the value of $48^{3}-30^{3}-18^{3}=48^{3}+(-30)^{3}+(-18)^{3}$.

Here, $48+(-30)+(-18)=0$

So, $48^{3}+(-30)^{3}+(-18)^{3}=3 \times 48 \times(-30) \times(-18)=77760$