Question

PLEASE guys answer this question quickly........​

Answer 1

Step-by-step explanation:

Use: cos(A+B) = cosAcosB-sinAsinB

$=cosx +cos(120 +x) +cos(x-120)\\=cosx + cos(120)cos(x)-sin(120)sin(x) + cos(x)cos(120)+sin(x)sin(120)\\=cosx-\frac{1}{2}cosx -\frac{1}{2}cosx=0$

Next use: a^3 + b^3 + c^3 = (a + b + c) (a^2+ b^2+ c^2– ab – bc – ca) + 3abc

$=cos^3x+cos^3(x+120)+cos^3(x-120)\\= (cosx + cos(120+x)+cos(x-120))(...)+3(cosx)(cos(120+x))(cos(x-120))\\=0 +3(cosx)(cos(120+x))(cos(x-120))$

Next use cos(A+B)cos(A-B)=cos^2(A) - sin^2(B)

$=3(cosx)(cos^2x-sin^2(120))\\=3(cosx)(\frac{4cos^2x-3}{4})\\=\frac{3}{4}(4cos^3x-3cosx)\\=\frac{3}{4}cos(3x)$

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