Giải thích các bước giải:

Ta có:

$2(\cos^4x+\sin^4x+\cos^2x\sin^2x)^2-(\cos^8x+\sin^8x)$

$=2(\cos^4x+\sin^4x+2\cos^2x\sin^2x-\cos^2x\sin^2x)^2-(\cos^8x+\sin^8x)$

$=2((\cos^2x+\sin^2x)^2-\cos^2x\sin^2x)^2-(\cos^8x+\sin^8x)$

$=2(1^2-\cos^2x\sin^2x)^2-(\cos^8x+\sin^8x)$

$=2(1-\cos^2x\sin^2x)^2-(\cos^8x+\sin^8x)$

$=2(1-2\cos^2x\sin^2x+\cos^4x\sin^4x)-(\cos^8x+\sin^8x)$

$=2-4\cos^2x\sin^2x-(\cos^8x\sin^8x-2\cos^4x\sin^4x)$

$=2-4\cos^2x\sin^2x-(\cos^4x-\sin^4x)^2$

$=2-\sin^22x-(\cos^2x-\sin^2x)^2(\cos^2x+\cos^2x)^2$

$=2-\sin^22x-\cos^22x\cdot 1^2$

$=2-\sin^22x-\cos^22x$

$=2-(\sin^22x+\cos^22x)$

$=2-1$

$=1$