Prove that `3(sinx-cosx)^4+4(sin^6x+cos^6x)+6(sinx+cosx)^2=13`

Explanation

`4(sin^6x+cos^6x)`

`=4[(sin^2x)^3+(cos^2x)^3]`

`=4[(sin^2x+cos^2x)(sin^4x−sin^2xcos^2x+cos^4x)]`

`=4[(sin^2x+cos^2x)^2−2sin^2xcos^2x−2sin^2xcos^2x]`

`=4[1−3sin^2xcos^2x]`

`=4−12sin^2xcos^2x ........(1)`

Now,

`6[sinx+cosx]^2`

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