Ta có $(\sin x+\cos x)^2=m^2\Leftrightarrow 1+2\sin x.\cos x=m^2\Leftrightarrow \sin x.\cos x=\frac{m^2-1}{2}$
a)$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x.\cos^2x$$=1-2\left(\frac{m^2-1}2 \right)^2=1-\frac{(m^2-1)^2}{2}=\frac{-m^4+2m^2+1}{2}$
b) $\sin^3x+\cos^3x=(\sin x+\cos x)^3-3\sin x\cos x(\sin x+\cos x)=m^3-\frac{3m(m^2-1)}{2}$
$=\frac{-m^3+3m}{2}$
c) $\sin^6 x +\cos^6 x=(\sin^3x+\cos^3x)^2-2\sin^3x\cos^3x$
$=\left( \frac{-m^3+3m}2 \right)^2-2\left( \frac{m^2-1}2 \right)^3$
d)$\sin^8x+\cos ^8x=(\sin^4x+\cos^4x)^2-2\sin^4x\cos^4x$
$=\left( \frac{-m^4+2m^2+1}2 \right)^2-2\left(\frac{m^2-1}2 \right)^4$