Ta có
$\sin^8x + \cos^8x = (\sin^4x + \cos^4x)^2 - 2\sin^4x \cos^4x$
$= [(\sin^2x + \cos^2x)^2 - 2\sin^2 \cos^2 ]^2 - 2\sin^4x \cos^4x$
$= (1 - 2\sin^2x \cos^2x)^2 - 2\sin^4x \cos^4x$
$= 1 + 4\sin^4x \cos^4x - 4\sin^2x \cos^2x - 2\sin^4x \cos^4x$
$= 1 + 2\sin^4x \cos^4x - 4\sin^2x \cos^2x= VP$