If in a  $\Delta ABC$ , ${\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C=3\cos A\cos B\cos C$  then the triangle is 

$a^3+b^3+c^3=3abc\Rightarrow a+b+c=0\left(or\right)a=b=c$

$\therefore {\cos }^{3}A+{\cos }^{3}B+{\cos }^{3}C=3\cos A\cos B\cos C$

$\Rightarrow \cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\ne 0$

$\Rightarrow \cos A=\cos B=\cos C$

$\Rightarrow A=B=C(\because 0<A,B,C<\pi )$

$\Rightarrow \Delta ABC$is equilateral