\(\left(x+y+z\right)^3-x^3-y^3-z^3\\ =x^3+y^3+z^3-x^3-y^3-z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\\ =3\left(x+y\right)\left(y+z\right)\left(z+x\right)\:\left(đpcm\right)\)
• \(VT=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3z+\left(x+y\right)^2+3xz^2+3yz^2-x^3-y^3\)
\(=3x^2y+3xy^2+3z\left(x^2+2xy+y^2\right)+3xz^2+3yz^2\)
\(=3x^2y+3xy^2+3x^2z+6xyz+3y^2z+3xz^2+3yz^2\) (1)
• \(VP=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3x+3y\right)\left(y+z\right)\left(z+x\right)\)
\(=\left(3xy+3xz+3y^2+3yz\right)\left(z+x\right)\)
\(=3xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2+3xyz\)
\(=6xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2\) (2)
Từ (1) và (2) suy ra \(VT=VP\) (đpcm)