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prove that `cos^3(2x)+3cos2x=4(cos^6x-sin^6x)`

Answer» `L.H.S. = cos^3(2x) +3cos2x`
`=cos2x(cos^2 (2x) +3)`
`=cos2x((2cos^2x-1)^2 +3)`
`=cos2x((4cos^4x+1 - 4cos^2x) +3)`
`=cos2x(4cos^4x+4 - 4cos^2x)`
`=4(2cos^2x -1)(cos^4x+1 - cos^2x)`
`=4(2cos^6x + 2cos^2x -2cos^4x - cos^4x- 1 + cos^2x)`
`=4(2cos^6x-3cos^4x+3cos^2x-1)`
`=4(2cos^6x+3cos^2x(1-cos^2x) - 1)`
`=4(2cos^6x+3cos^2xsin^2x - (sin^2x+cos^2x)^3)`
`=4(2cos^6x+3cos^2xsin^2x - sin^6x-cos^6x - 3sin^2xcos^2x (sin^2x+cos^2x))`
`=4(cos^6x - sin^6x+3cos^2xsin^2x-3cos^2xsin^2x)`
`=4(cos^6x - sin^6x) = R.H.S.`


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