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Prove that `cos^3A+cos^3(120^0+A)+cos^3(240^0+A)=3/4cos3A`

Answer» `cosA + cos(120^@+A) +cos(240^@+A)`
`= cosA + (2cos((120+A+240+A)/2)cos((120+A-240-A)/2))`
`= cosA + (2cos(180+A)cos(-60^@))`
`= cosA + (2cos(180+A)cos60^@)`
`= cosA + (-2cosA(1/2))`
`=cosA - cosA = 0`
We know, when, `a+b+c = 0`,
`a^3+b^3+c^3 = 3abc`
`:. L.H.S. = cos^3A + cos^3(120^@+A) +cos^3(240^@+A) = 3cosAcos(120^@+A)cos(240^@+A)`
`=3cosAcos(180-(60-A))cos(180+(60+A))`
`=3cosAcos(-(60-A))cos(-(60+A))`
`=3cosAcos(60-A)cos(60+A)`
Using `cos(C-D)cos(C+D) = cos^2C - sin^2D`
`=3cosA[cos^2 60^@-sin^2A]`
`=3cosA[1/4-(1-cos^2A)]`
`=3cosA[-3/4+cos^2A]`
`=3/4(4cos^3A - 3cosA)`
`=3/4cos3A = R.H.S.`


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