Question

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

Answer 1

We have,

${\cos }^{3}x+{\cos }^3(120-x)+{\cos }^3(120+x)$

Now, Using

$\cos 3x=4{\cos }^{3}x-3\cos x$

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

Such that,

$\frac{\cos 3x+3\cos x}{4}+\frac{\cos (360-3x)+3\cos (120-x)}{4}+\frac{\cos (360+3x)+3\cos (120+x)}{4}$

$=\frac{1}{4}[\cos 3x+3\cos x+\cos (360-3x)+3\cos (120-x)+\cos (360+3x)+3\cos (120+x)]$

$=\frac{3}{4}[\cos 3x+\cos x+\cos (120-x)+\cos (120+x)]$

$=\frac{3}{4}[2\cos 2xcosx+2cos120cosx]$

$=\frac{3}{4}[2\cos 2xcosx+2(-\frac{1}{2})cosx]$

$=\frac{3}{4}[\cos 3+\cos x-\cos x]$

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

Minimum value of $\cos 3x=-1$

So, minimum value of ${\cos }^{3}x+{\cos }^3(120-x)+{\cos }^3(120+x)=\frac{3}{4}\times -1=\frac{-3}{4}$

Hence, this is the answer.