Question

`x^3+y^3 +z^3-3xyz` Phân tích 

Answer 1

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

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