Prove that :sin^6A+cos^6A=1-3sin^2Acos^2A
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sin⁶A + Cos⁶A = (sin²A)³ + (cos²A)³
using a³ + b³ = (a + b)³ - 3ab(a+b)
we get
(sin²A + cos²A)³ - 3sin²Acos²A(sin²A + cos²A)
we know sin²A + Cos²A = 1
hence 1 - 3sin²Acos²A proved
using a³ + b³ = (a + b)³ - 3ab(a+b)
we get
(sin²A + cos²A)³ - 3sin²Acos²A(sin²A + cos²A)
we know sin²A + Cos²A = 1
hence 1 - 3sin²Acos²A proved
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