The minimum value of ${\cos }^{3}x+{\cos }^3\left(120°+x\right)+{\cos }^3\left(120-x\right)$  is

If $a+b+c=0$ , then $a^3+b^3+c^3=3abc$

${\cos }^{3}x+{\cos }^3\left(120+x\right)+{\cos }^3\left(120-x\right)=3.\cos x.\cos \left(120+x\right).\cos \left(120-x\right)$

$=3\left(\frac{1}{4}\cos 3x\right)$

$=\frac{3}{4}\cos 3x$

We have $-1\le \cos 3x\le 1$

$-\frac{3}{4}\le \frac{3}{4}\cos 3x\le \frac{3}{4}$