Show That `2(sin^6x+cos^6x)-3(sin^4x+cos^4x)+1=0`

Explanation

`L.H.S. = 2(sin^6x+cos^6x) - 3(sin^4x+cos^4x) +1`

`= 2((sin^2x)^3+(cos^2x)^3) - 3((sin^2x)^2+(cos^2x)^2) +1`

As, `a^3+b^3 = (a+b)^3-3ab(a+b) and a^2+b^2 = (a+b)^2-2ab`,

So, it becomes,

`= 2((sin^2x+cos^2x)^3 - 3sin^2xcos^2x(sin^2x+cos^2x)) - 3((sin^2x+cos^2x)^2 - 2sin^2xcos^2x) +1`

`=2(1-3sin^2xcos^2x) - 3(1-2sin^2xcos^2x)+1`

`=2-6sin^2xcos^2x-3+6sin^2xcos^2x+1`

`=0 = R.H.S.`