Question

Rút gọn \(\sin^6x+\cos^6x+3\sin^2x\cos^2x\)

Answer 1

\(sin^6x+cos^6x+3sin^2x.cos^2x=\left(sin^2x\right)^3+\left(cos^2x\right)^3+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left[\left(sin^2x\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2\right]+3sin^2x.cos^2x\)

\(=1.\left[\left(sin^2\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2\right]+3sin^2x.cos^2x\)

\(=\left(sin^2x\right)^2-sin^2x.cos^2x+\left(cos^2x\right)^2+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2=1^2=1\)