Ta có
$\sin^4x + \cos^4x - 2(\sin^6x + \cos^6x) = (\sin^2x + \cos^2x)^2 - 2\sin^2x \cos^2x - 2(\sin^2x + \cos^2x)(\sin^4x + \cos^4x - \sin^2x \cos^2x)$
$= 1 - 2\sin^2x \cos^2x - 2(\sin^4x + \cos^4x) + 2\sin^2x \cos^2x$
$=1 - 2[(\sin^2x + \cos^2x)^2 - 2\sin^2x \cos^2x]$
$=1 - 2(1 - 2\sin^2x \cos^2x)$
$= 4\sin^2x \cos^2x - 1$
$= \sin^2(2x) - 1$
Vậy đẳng thức đã cho vẫn phụ thuộc vào $x$.