\(z^3\left(x+y^2\right)+y^3\left(z-x^2\right)-x^3\left(y+z^2\right)-xyz\left(xyz-1\right)\)

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

\(=\left(y^2z^3+y^3z\right)+\left(xz^3+xyz\right)-\left(x^2y^3+x^2y^2z^2\right)-x^3\left(y+z^2\right)\)

\(=y^2z\left(y+z^2\right)+xz\left(y+z^2\right)-x^2y^2\left(y+z^2\right)-x^3\left(y+z^2\right)\)

\(=\left(y+z^2\right)\left(y^2z+xz-x^2y^2-x^3\right)\)

\(=\left(y+z^2\right)\left[z\left(y^2+x\right)-x^2\left(y^2+x\right)\right]\)

\(=\left(y+z^2\right)\left(z-x^2\right)\left(y^2+x\right)\)

Tick hộ nha bạn 😘

z^3(x+y^2)+y^3(z-x^2)-x^3(y+z^2)-xyz(xyz-1)

1D  2C

Câu 1: D

Câu 2: C

Đặt \(\left\{{}\begin{matrix}a=x+y\\b=y+z\\c=x+z\end{matrix}\right.\Leftrightarrow x+y+z=\dfrac{a+b+c}{2}\)

\(8\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(z+x\right)^3\\ =8\left(\dfrac{a+b+c}{2}\right)^3-a^3-b^3-c^3\\ =\left(a+b+c\right)^3-a^3-b^3-c^3\\ =\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-\left(a+b\right)^3+3ab\left(a+b\right)-c^3\\ =3\left(a+b\right)\left(ac+bc+c^2+ab\right)\\ =3\left(a+b\right)\left(b+c\right)\left(a+c\right)\\ =3\left(x+y+y+z\right)\left(y+z+z+x\right)\left(z+x+x+y\right)\\ =3\left(x+2y+z\right)\left(x+y+2z\right)\left(2x+y+z\right)\)

Hướng dẫn

Đặt là x,y,z

Chứng minh được là \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(\left(x+y+z\right)^3=x^3+y^3+z^3+\left(x+y\right)\left(x+z\right)\left(y+z\right)\)

\(\left(x+y+z\right)^3-x^3+y^3+z^3\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)