Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)

Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)

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

\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Hay \(A=3\cdot2x\cdot2y\cdot2z\)

\(A=24xyz\)

Đặt y-z=-[(x-y)+(z-x)]

Thay vào rồi cm nha bạn

Ta có :

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

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

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

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

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

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

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

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

Vậy ...

CÓ AI CHƠI GAME BANG BANG K NÀO

Ta có

(x+y+z)³-x³-y³-z³

=x³+y³+z³+3(x+y)(x+z)(z+y)-(x³+y³+z³)

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

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

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

\(=2xy+2xz+2yz\)

\(=2\left(xy+xz+yz\right)\)

Đc chưa ?

Phương Đỗ Sai rùi bạn.

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

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

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

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

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

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

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

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)\(=3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[\left(xy+xz\right)+\left(yz+z^2\right)\right]\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
P/s: Bài viết sử dụng 1 công thức đặc biệt:  (a+b)^3 = a^3 + b^3 + 3ab(a+b)
Công thức được suy ra từ hàng đẳng thức số 4 khi:
(a+b)^3 = a^3 + 3*a^2*b + 3*a*b^2 + b^3  = a^3 + 3*a*b*(a+b) + b^3