How to prove: #sin^6theta-cos^6theta=(2sin^2theta-1)(cos^2theta+sin^4theta)# ?

1 Answer
Abhishek K.
Jun 4, 2018

#LHS=sin^6x-cos^6x#

#=(sin^2x)^3-(cos^2x)^3#

#=[sin^2x-cos^2x]*[(sin^2x)^2+sin^2x*cos^2x+(cos^2x)^2]#

#=[sin^2x-(1-sin^2x)][sin^4x+sin^2x*cos^2x+cos^4x]#

#=[sin^2x-1+sin^2x][sin^4x+cos^2x(sin^2x+cos^2x)]#

#=[2sin^2x-1][sin^4x+cos^2x]=RHS#

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