10sin'2x -2sinxcosx= 8cos'2x => 10sin'2x-2sinxcosx-8cos'2x=0 => 5sinx'2 - sinxcosx - 4cos'2x=0 => (5sinx+4cosx)*(sinx-cosx)=0 1) sinx= -4cosx 2) sinx=cosx Sinx/cosx=-4 => tgx =-4/5 => x = arctg(-4/5)
Решение задания приложено
=a³-2a²+2²a+2*(a²-2a+2²)=a³-2a²+4a+2a²-4a+8=a³+8=a³+2³.