Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3x^2yz+3xy^2z+3xyz^2-x^3-y^3-z^3\)
\(=3x^2yz+3xy^2z+3xyz^2\)
\(=3xyz\left(x+y+z\right)\)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y^3+z^3\right)\)
\(=\left(y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz+x^2+xy+xz+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)
\(=\left(y+z\right)\left[\left(3x^2+y^2+z^2+3xy+3xz+2yz\right)-\left(y^2-yz+z^2\right)\right]\)
\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz-y^2+yz-z^2\right)\)
\(=\left(y+z\right)\left(3x^2+3xy+3xz+3yz\right)\)
\(=\left(y+z\right)\left[3x\left(x+y\right)+3z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(3x+3z\right)\)
\(=3\left(x+z\right)\left(y+z\right)\left(x+y\right)\)
( x +y + z)3 - x3 - y3 - z3
= ( x + y + z - x).[( x + y +z)2 + ( x+y+z).x + x2] - ( y3 + z3)
= ( y +z).[( x + y +z)2 + ( x+y+z).x + x2] -( y +z).( y2 - zy + z2)
= ( y + z).[( x + y +z)2 + ( x+y+z).x + x2 - y2 + yz - z2]
Còn lại cậu tính như thường thôi nha !( Dài chết xỉu 
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