Ý bạn là phân tích đa thức thành nhân tử hả.

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

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

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

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

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

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

Ta có: 

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

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

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

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

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

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

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

 x^3+y^3+z^3-3xyz 
=(x+y+z)^3-3x^2.y-3x.y^2-3y^2.z-3y.z^2... 
=(x+y+z)^3-3xy(x+y+z)-3yz(x+y+z)-3xz(x... 
=(x+y+z)(<x+y+z>^2-3xy-3yz-3xz) 
=(x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xy-3... 
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)                      

Biến đổi tương đương nhé bạn.

a: Ta có: \(\left(x+y\right)^2\)

\(=x^2+2xy+y^2\)

\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)

a,(4x+1)³-(x-2)³

=(4x+1-x+2). \(\left[\left(4x+1\right)^2+\left(4x+1\right)\left(x-2\right)+\left(x-2\right)^2\right]\)

=(3x+3).(16x²+8x+1+4x²-8x+x-2+x²-4x+4)

=3(x+1).(21x²-3x+3)=3(x+1).3(7x²-x+1)

=9.(x+1)(7x²-x+1)

đề câu 2 bạn ghi sai rồi

x³+y³+z³-3xyz=x³+3x²y+3xy²+y³-3xyz-3x²y-3xy²+z³

=(x+y)³+z³-3xy(x+y+z)

=(x+y+z). ((x+y)²-(x+y)z+z²)-3xy(x+y+z)

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

=(x+y+z)(x²+y²+z²-xz-yz+2xy-3xy)

=(x+y+z)(x²+y²+z²-xz-yz- xy)

Ta có \(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\right]=0\)(Nhân hai vế với 2)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)

Tới đây bạn xét hai trường hợp nhé :)

(x+y+z)((X+Y)^2-Z(X+Y))-3XY(X+Y+Z)

=(X+Y+Z)(X^2+2XY+Y^2-XZ-YZ-3XY)

=(X+Y+Z)(X^2+Y^2+Z^2-XZ-YZ-XY)

x^3+y^3+z^3-3xy(x+y+z)=0

x^3+y^3+z^3-3xy*xyz=0

3xyz-3xyz=0

chuc ban thanh cong

\(\left(a+b\right)^3-\left(a-b\right)^3\)

\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)

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

\(=6a^2b+2b^3\)

\(=2b\left(3a^2+b^2\right)\)

a/\(\left(a+b\right)^3-\left(a-b\right)^3\)

\(=\left(a^3+3a^2b+3ab^2+b^3\right)-\left(a^3-3a^2b+3ab^2-b^3\right)\)\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^2\)

\(=6ab^2+2b^3\)(rút gọn hết)

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

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

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

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

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

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

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

Hok tốt

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

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

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

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

a, \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

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

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

\(\Leftrightarrow x^3+y^3+z^3=3xyz\)(vì x+y=-z)

Cảm ơn ạ