a,
$=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)$
$=\sin^4x+\cos^4x-\sin^2x\cos^2x$
$=1-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.4\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.\sin^22x$
$=1-\dfrac{3}{4}.\dfrac{1-\cos 4x}{2}$
$=1-\dfrac{3-3\cos 4x}{8}$
$=\dfrac{5}{8}+\dfrac{3}{8}\cos 4x$
b,
$=(\sin^4x)^2+(\cos^4x)^2$
$=(\sin^4x+\cos^4x)^2-2\sin^4x\cos^4x$
$=(1-2\sin^2x\cos^2x)^2-2\sin^4x\cos^4x$
$=1-4\sin^2x\cos^2x+4\sin^4x\cos^4x-2\sin^4x\cos^4x$
$=1-\sin^22x+2\sin^4x\cos^4x$
$=\cos^22x+2\sin^4x\cos^4x$
$=\dfrac{1-\cos4x}{2}+2(\dfrac{1}{2}\sin 2x)^4$
$=\dfrac{1-\cos4x}{2}+\dfrac{1}{8}\sin^42x$
$=\dfrac{1}{2}-\dfrac{1}{2}\cos 4x+\dfrac{1}{8}.\dfrac{(1-\cos4x)^2}{4}$
