(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...

Approved