Question

prove that :
Sin^6 A + cos^6 A = 1 - 3sin^2 A *cos^2 A​

Answer 1

sin^2A + cos^2A =1

Cubing on both sides

(sin^2A+cos^2A)^3=1^3

(sin^2A)^³+(cos^2A)^3 + 3sin^2Acos^2A(sin^2A+cos^2A)=1

sin^6A+cos^6A+3sin^2Acos^2A(1)=1

sin^6A+cos^6A=1-3sin^2Acos^2A

Hence, proved.