`~rai~`
\(\sin^6x+\cos^6x\\=(\sin^2x)^3+(\cos^2x)^3\\=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\=\sin^4x-\sin^2\cos^2x+\cos^4x\quad\text{(do }\sin^2x+\cos^2x=1)\\=(\sin^4x+2\sin^2x\cos^2x+\cos^4x)-3\sin^2x\cos^2x\\=(\sin^2x+\cos^2x)^2-\dfrac{3}{4}.4\sin^2x\cos^2x\\=1-\dfrac{3}{4}\sin^22x.\)