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(sinx)3-(cosx)3=[(sinx)2-(cosx)2][1-2*(sinx)2*(cosx)2]
here (sinx)3 means cube of sinx and similarly other powers .
(sinx)3 -(cosx)3 = (sinx-cosx)(sin2 x +cos2 x -sinxcosx) or (sinx+cosx)(1-sin2x/2)
sin2 x -cos2 x =(sinx-cosx)(sinx+cosx)
on putting these
we get
(sinx+cosx)(1-sin2x/2 ) = ((sinx+cosx)(sinx-cosx)(1-2(sinxcosx)2 ) or
(sinx+cosx) {1-sin2x/2 - (sinx-cosx)(1-2(sinxcosx)2 } = 0
sinx +cosx =0 or {1-sin2x/2 -(sinx-cosx)(1-2(sinxcosx)2 }=0
tanx=-1 or {1-sin2x/2 -(sinx-cosx)(1-(sin2x)2 /2 ) =0
x =npi -pi/4 or (sinx-cosx)2 =1-sin2x ,so nw sustitute the value of sin2x in above eq
& simplify it...
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