用公式a³+b³=(a+b) (a²-ab+b²)
cos^6x+sin^6x
=(cos²x)³ + (sin²x)³
=(cos²x+sin²x)[(cos²x)² - cos²x * sin²x + (sin²x)²]
=(1-sin²x)² - (1-sin²x) * sin²x + sin^4x
=1-2sin²x + sin^4x -sin²x + sin^4x + sin^4x
=1-3sin²x +3sin^4x
三角等式求证：cos^6x+sin^6x=1-3sin^2x+3sin^4x
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