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

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

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

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

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

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

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

x³ + y³ + z³ - 3xyz

= (x+y)³ - 3xy(x-y) + z³ - 3xyz 
= [(x+y)³ + z³] - 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)

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)

  Ta có: 
x³ + y³ + z³ - 3xyz = (x+y)³ - 3xy(x-y) + z³ - 3xyz 
= [(x+y)³ + z³] - 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). 

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

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

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

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

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

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

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

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

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

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

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

Trần Đức Thắng sai rùi X^3+y^3+z^3+3xyz cơ mà có phải X^3+y^3+z^3-3xyz đâu mà làm vậy 

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

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

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

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