Question

X^3+y^3+z^3-(x+y+z)^3
разложить на множители

Answer 1

-3(x+y)(z+y)(z+x)

Пошаговое объяснение:

(x+y)(x^2+y^2-xy)-(x+y)(z^2+(x+y+z)^2+z*(x+y+z))=

=(x+y)(x^2+y^2-xy)-(x+y)(z^2+x^2+y^2+z^2+xz+yz+z^2+2xy+2xz+2yz)=

=(x+y)(x^2+y^2-xy)-(x+y)(3z^2+x^2+y^2+3xz+3yz+2xy)=

=(x+y)(x^2+y^2-xy-3z^2-x^2-y^2-3xz-3yz-2xy)=

=(x+y)(-3xy-3z^2-3xz-3yz)=-3(x+y)(z^2+xy+xz+yz)=

=-3(x+y)(z(z+y)+x(z+y))=-3(x+y)(z+y)(z+x)

Answer 2

$-3(x+y)(x+z)(y+z)$

Пошаговое объяснение:

Выполним преобразования:

1)

$(x+y+z)^3-z^3=(x+y+z-z)((x+y+z)^2+(x+y+z)z+z^2)=\\ =(x+y)((x+y+z-z)^2+3(x+y+z)z)=(x+y)((x+y)^2+3(x+y+z)z)$

2)

$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)$

Подставляем в условие:

3)

$x^3+y^3+z^3-(x+y+z)^3=(x+y)((x+y)^2-3xy)-(x+y)((x+y)^2+3(x+y+z)z)=\\ =(x+y)((x+y)^2-3xy-(x+y)^2-3(x+y+z)z)=(x+y)(-3xy-3(x+y+z)z)=\\ =-3(x+y)(xy+xz+(y+z)z)=-3(x+y)(x(y+z)+(y+z)z)=\\ =-3(x+y)(x+z)(y+z)$