Question

2(sin6A+cos6A)-3(sin4A+cos4A)+1=0

Answer 1

2(sin6a+cos6a)

-3(sin4a+cos4a)+1=0

=2[(sin2a)3+(cos2a)3]

-3[(sin2a)2+(cos2a)2]+1

=2[(sin2a+cos2a)3

-3sin2acos2a(sin2a+cos2a)]

-3[(sin2a+cos2a)2

-2sin2acos2a]+1

=2[1-3sin2acos2a]

-3[1-2sin2acos2a]+1

=2-6sin2acos2a

-3+6sin2acos2a+1
=0

Hence proved
 Tabbu Singh answered this
2(sin6a+cos6a)

-3(sin4a+cos4a)+1=0

=2[(sin2a)3+(cos2a)3]

-3[(sin2a)2+(cos2a)2]+1

=2[(sin2a+cos2a)3

-3sin2acos2a(sin2a+cos2a)]

-3[(sin2a+cos2a)2

-2sin2acos2a]+1

=2[1-3sin2acos2a]

-3[1-2sin2acos2a]+1

=2-6sin2acos2a

-3+6sin2acos2a+1=0

Hence prove Thank you