Sửa đề: \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
x3 + y3 + z3 - 3xyz
= (x³ + 3x²y + 3xy² + y³) - (3x²y - 3xy²) + z³ - 3xyz
= (x + y)³ - 3xy(x - y) + z³ - 3xyz
= [(x + y)³ + z³] - 3xy(x + y + z)
= (x + y + z)³ - 3(x + y)²z - 3(x + y)z² - 3xy(x + y + z)
= (x + y + z)³ - 3z(x + y)(x + y + z) - 3xy(x + y + z)
= (x + y + z)[(x + y + z)² - 3z(x + y) - 3xy]
= (x + y + z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x + y + z)(x² + y² + z² - xy - xz - yz)