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

Explanation

`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)`

...