X^3+y^3+z^3-3*x*y*z-(1/2)(x+y+-y)^2+(x-z)^2+(y-z)^2) - вопрос №3079197333312222 от violakim2806 27.02.2022 19:52

Question

Алгебра: X^3+y^3+z^3-3*x*y*z-(1/2)(x+y+-y)^2+(x-z)^2+(y-z)^2) решить

violakim2806 · Answer

Преобразуем его  так:(x^3-xyz) +(y^3-xyz) +(z^3-xyz)-1/2(x+y+z)(2x^2+2y^2+2z^2-2xy-2xz-2yz)x*(x^2-yz) +y*(y^2-xz)+z*(z^2-xy)-(x+y+z)( (x^2-yz) +(y^2-xz) +(z^2-xy) )=(y+z)*(yz-x^2) +(x+z)(xz-y^2) +(x+y)*(xy-z^2)=y^2*z -y*x^2+z^2*y-z*x^2 +x^2*z -y^2*x+z^2*x-y^2*z +x^2*y -z^2*x+y^2*x -z^2*y=0ответ:0...