$y=\sin^6x+\cos^6x$
$=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x$
$=1.[(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x-\sin^2x\cos^2x]$
$=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.4\sin^2x\cos^2x$
$=1-\dfrac{3}{4}\sin^22x$
$=1-\dfrac{3}{4}\Big(\dfrac{1}{2}-\dfrac{1}{2}\cos4x\Big)$
$=\dfrac{3}{8}\cos4x+\dfrac{5}{8}$
Ta có: $-1\le \sin4x\le 1$
$\Leftrightarrow \dfrac{1}{4}\le y\le 1$
Vậy: $m=\dfrac{1}{4}; M=1$
