sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）

原式=sin^8X-2sin^4xcos^4X+cos^8X+4sin^2Xcos^2x
=（sin^4x-cos^4x）^2+4sin^2Xcos^2x
=[（sin^2x-cos^2x）（sin^2x+cos^2x）]^2+4sin^2Xcos^2x
=（sin^2x-cos^2x）^2+4sin^2Xcos^2x
=sin^4x-2sin^2Xcos^2x+cos^4x+4sin^2Xcos^2x
=sin^4x+2sin^2Xcos^2x+cos^4x
=（sin^2x+cos^2x）^2
=1

答案为1
化简过程：=sin^8X-sin^4xcos^4X+cos^8X-sin^4xcos^4X+4sin^2Xcos^2x
=sin^4X（sin^4X-cos^4X）+cos^4X（cos^4X-sin^4x）+4sin^2Xcos^2x
=sin^4X（sin^2X-cos^2X））+cos^4X...