cos⁴A-sin⁴A=x

find cos⁶-sin⁶ in terms of x

Dear ajinkya

cos⁴A-sin⁴A=x

(cos²A-sin²A)(cos²A+sin²A)=x

cos²A-sin²A =x

cos2A =x

I= cos⁶A-sin⁶A =(cos²A)³-(sin²A)³

                         =(cos²A-sin²A)[(cos²A)²+(sin²A)² +(cos²A)(sin²A)]

                         =x[(cos²A-sin²A)² + 3(cos²A)(sin²A)]

                           =x[x² +3/4 (sin2A)²]

                           =x[x² +3/4 (1-(cos2A)²)]

                            =x[x² +3/4 (1-x²)]

                            =x [3+x²]/4

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