a) =x³+3x²y+3xy²+y³-x³+3x²y-3xy²+y³

=6x²y+2y³

=2y(3x²+y)

  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). 

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

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

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

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

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

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

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

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 

\(\left(3x+1\right)^2-\left(3x-1\right)^2\)

\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

\(\left(x+y\right)^3+\left(x-y\right)^3\)

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

\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\cdot\left(x^2+3y^2\right)\)

______

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

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

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

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

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\left(1\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)\)

\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\)

có tkế dừng lại ở (1) cũg đk

cảm ơn bạn nhé

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

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

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

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

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

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

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

                                             \(=\left(x+y+z\right).\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy.\left(x+y+z\right)\)

                                               \(=\left(x+y+z\right).\left(x^2+y^2+z^2-zx-zy+2zy-3xy\right)\)

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

Vừa làm xong . Chúc bạn học tốt !

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

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

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

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

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). 
~~~~~~~~ 
Bài làm trên mình đã sử dụng hằng đẳng thức đáng nhớ sau: 
(a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a-b) 
=> a³ + b³ = (a+b)³ - 3ab(a-b).