prove that:- sin^6(A) + cos^6(A) = 1 - 3sin^2(A) cos^2(A)

sin⁶_x+cos⁶x=(sin²x)³+(cos²x)+3sin²xcos²x(sin²x+cos²x)-3sin²xcos²x(sin²x+cos²x)             =(sin²x+cos²x)³-3sin²xcos²x                           =1-3sin²xcos²x        

(A+B)^3=A^3+B^3+3AB(A+B)

=>A^3+B^3=(A+B)^3-3AB(A+B)

PUT A=SIN^2(A) AND B=COS^2(A)

SO A^3+B^3  WILL BE  SIN^6(A)+COS^6(A)

NOW SOLVE R.H.S. TO GET THE ANSWER