Решить(x-2y)^2(3x+4y)-x(3(x^2-4y^2)+8y(y-x))=

(х-2у)^2(3х+4у)-х(3(х2-4у^2)+8у(у-х))= (х-2у)^2×3х+(х-2у)^2×4у-х×(3х^2-12у^2+8у2-8ху)= (х^2-4ху+4у^2)×3х+(х-2у)^2×4у-х×(3х^2-4у^2-8ху)= 3х^3-12х^2у+12ху^2+(х-2у)^2×4у-3х^3+4ху2+8х^2у= -12х^2у+12ху^2+(х^2-4ху+4у^2)×4у+4ху^2+8^2у= -12х^2у+12ху^2+4х^2у-16ху^2+16у^3+4ху^2+8х^у= 16у^2