Question

phân tích đa thức thành nhân tử :x³+y³+z³-3xyz

= (x+y)³+z³-3xyz-3x²y-3xy² giải thích cho mik bước này vì ko hiểu lăm

Answer 1

\(x^3+y^3+z^3-3xyz\)

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

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

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

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

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

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

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

 

 

Answer 2

thui mik hieu roi cam on mn