請幫我解一題數學

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

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

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

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

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

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

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

x³ + y³ + z³ - 3xyz
= x³ + y³ + 3y²z + 3yz² + z³ - 3xyz - 3y²z - 3yz²
= x³ + (y³ + 3x²z + 3xz² + z³) - (3xyz + 3y²z + 3yz²)
= [x³ + (y + z)³] - 3yz(x + y + z)
= [x + (y + z)] [x² - x(y + z) + (y + z)²] - 3yz(x + y + z)
= (x + y + z)(x² - xy - xz + y² + 2yz + z²) - 3yz(x + y + z)
= (x + y + z)(x² + y² + z² - xy + 2yz - zx ) - 3yz(x + y + z)
= (x + y + z) [(x² + y² + z² - xy + 2yz - zx ) - 3yz]
= (x + y + z)(x² + y² + z² - xy - yz - zx )

x³+y³+(z³-3xyz )
=[( x+y)³-3x²y-3xy²]+(z³-3xyz)
=[(x+y)³+z³]-(3x²y+3xy²+3xyz)
=(x+y+z)[(x+y)²-(x+y)z+z²]-3xy(x+y+z)
=(x+y+z)(x²+y²+2xy-xz-yz+z²)-3xy(x+y+z)
=(x+y+z)(x²+y²+z²-xy-xz-yz)