prove the identity !!!!!!!!!!!!!!!!!!!!!!!

Question

cos³2A+3cos2A=4(cos^6A-sin^6A)

Answer 1

Pressumed that

4(cos⁶A - sin⁶A) = 4((cos²A)³ - (sin²A)³

                                  = 4(cos²A - sin²A)(cos⁴A + cos²Asin²A + sin⁴A)

                                 = 4(cos2A)(cos²A + sin²A)² - sin²Acos²A

                                 = 4cos2A(1 - sin²Acos²A)

                                = 4cos2A - 4cos2Asin²Acos²A

                                = 4cos2A - cos2Asin²(2A)

                               = 4cos2A - cos2A(1 - cos²(2A)

                               = 4cos2A - cos2A + cos³(2A)

                              = 3cos2A + cos³(2A)

                              = cos³(2A) + 3cos2A. 

Hence the prove the identity