Ta có:
\(sin^4a+cos^4a=\left(sin^2a+cos^2a\right)^2-2sin^2acos^2a=1-2sin^2a.cos^2a\)
Và:
\(sin^6a+cos^6a=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a.\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
Do đó:
\(A=3\left(1-2sin^2a.cos^2a\right)-2\left(1-3sin^2a.cos^2a\right)=1\)
\(B=1-3sin^2.cos^2a+3sin^2a.cos^2a=1\)