Ta có: M = xy(x+y) + yz(y+z) + xz (x+z) + 2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(x + y) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

M = (x + y)(y + z)(z + x) (đpcm)

( x + y + z )³ - x³ - y³ - z³

= [ ( x + y + z )³ - x³ ] - ( y³ + z³ )

= ( x + y + z - x )[ ( x + y + z )² + ( x + y + z )x + x² ] - ( y + z )( y² - yz + z² )

= ( y + z )( 3x² + y² + z² + 2yz + 3zx + 3xy ) - ( y + z )( y² - yz + z² )

= ( y + z )( 3x² + y² + z² + 2yz + 3zx + 3xy - y² + yz - z² )

= ( y + z )( 3x² + 3yz + 3zx + 3xy )

= 3( y + z )( x² + yz + zx + xy )

= 3( y + z )[ ( x² + zx ) + ( xy + yz ) ]

= 3( y + z )[ x( x + z ) + y( x + z ) ]

= 3( y + z )( x + z )( x + y )

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

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

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

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

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

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

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

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

\(=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]\)

\(=\left(y+z\right).\left(x+y\right).\left(3x+3z\right)\)

\(=3.\left(y+z\right).\left(x+y\right).\left(x+z\right)\)

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

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

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

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

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

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\left(x+y+z\right)^5-x^5-y^5-z^5\)

Xét phương trình: \(\left(x+y+z\right)^5-x^5-y^5-z^5=0\)

Có nghiệm: \(x=-y;x=-z;y=-z\)

Hệ số của mũ là: 5

\(\Rightarrow\left(x+y+z\right)^5-x^5-y^5-z^5\)

\(=5\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(x^2+y^2+z^2+xy+yz+xz\right)\)

Hok Tốt!!!

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

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

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

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

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

Giỏi toán cần phải cọ xát nhiếu;

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

Bạn thêm vào 2 hạng tử , sau đó bớt 2 hạng tử để biểu thức ko thay đổi nhé, ở đây xuất hiện 1 hằng đẳng thức:

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

Ta thấy lại tiếp tục xuất hiên 1 hằng đẳng thức: a^3+b^3 nên ta có:

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

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

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

ủng hộ nha các bạn

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

=(x+y)^3+3(x+y)^2z +3(x+y)z^2+z^3-x^3-y^3-z^3

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

=3xy(x+y)+3(x+y)^2z+3(x+y)z^2

=3(x+y)(xy+(x+y)z+z^2)

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

=3(x+y)(x(y+z)+z(y+z))

=3(x+y)(y+z)(x+z)

1) a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3

= -3(a^2b-ab^2+b^2c-bc^2+c^2a-ca^2)

=-3(ab(a-b)+c(b^2-a^2)-c^2(b-a))

= -3(ab(a-b)-c(a+b)(a-b)+c^2(a-b))

= -3(a-b)(ab-ac-bc+c^2)

= -3(a-b)(a(b-c)-c(b-c))

= -3(a-b)(b-c)(a-c)

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

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

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

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

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

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