$\displaystyle if sinA+cosA=x, Prove that Sin^6A +cos^6A = 4-3(x^2-1)^2 /4 $
something is wrong here... on the left side you have sin( A ) and on the right side x ???
Anyway... if A = x...
sin^6( x ) + cos^6( x ) = ( sin^2( x ) + cos^2( x ) )^3 - 3sin^4( x )*cos^2( x ) - 3sin^2( x )*cos^4( x ) =
= 1^3 - 3*( sin^2( 2x )/4 )*sin^2( x ) - 3( sin^2( 2x )/4 )*cos^2( x ) =
= 1 - (3/4)*sin^2( 2x )*( sin^2( x ) + cos^2( x ) ) =
= 1 - (3/4)*sin^2( 2x )