Sai bạn nhé.
$\sin^2a+\cos^2a=1$
$\Rightarrow (\sin^2a+\cos^2a)^2=1$
$\Leftrightarrow (\sin^2a)^2+(\cos^2a)^2=1-2\sin^2a\cos^2a\neq 1$
Phân tích $\sin^6a+\cos^6a$ như sau:
$\sin^6a+\cos^6a$
$=(\sin^2a)^3+(\cos^2a)^3$
$=(\sin^2a+\cos^2a)(\sin^4a-\sin^2a\cos^2a+\cos^4a)$
$=(\sin^2a)^2+(\cos^2a)^2-\sin^2a\cos^2a$
$=1-3\sin^2a\cos^2a$ (vì $(\sin^2a+\cos^2a)^2=1$)