Câu hỏi của Phan Việt Hoàng

Question

Phân tích đa thức thành nhân tử :

x^3 + y^3 + z^3 - 3 xyz

nguyễn thị huyền anh · Accepted Answer

$=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz$

$=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz$

$=\left(x+y+z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)$

$=\left(x+y+z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2-3xy\right]$

$=\left(x+y+z\right)\left(x^2+y^2+z^2+xz+yz-xy\right)$

Câu trả lời:

$=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz$

$=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz$

$=\left(x+y+z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)$

$=\left(x+y+z\right)\left[\left(x+y\right)^2+z\left(x+y\right)+z^2-3xy\right]$

$=\left(x+y+z\right)\left(x^2+y^2+z^2+xz+yz-xy\right)$

Đường Quỳnh Giang · Answer

$x^3+y^3+z^3-3xyz$

$=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz$

$=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)$

$=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$

Câu trả lời:

$x^3+y^3+z^3-3xyz$

$=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz$

$=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)$

$=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$